



LIBRA RY OF CONG RESS. 

Chap._._._ (Copyright No..„:„__. 



Shell 



:Ei 



UNITED STATES OF AMERICA. 




J. T. Fairchild. 



COMPLETE AND PRACTICAL 



SOLUTION BOOK 



FOR THE 



Common School Teacher 



BY 

Ji T. FAIRCHILD, A. M., Ph. M. 

INSTRUCTOR IN MATHEMATICS, ENGLISH GRAMMAR, LATIN, 

CHEMISTRY, PHYSICS AND BOTANY, CRAWFIS 

COLLEGE, CRAWFIS COLLEGE, O. 




Published by the Author 
1899 



31674 

COPYRIGHT, 1898, 

BY 

J. T. Fairchild. 
TWO eo^es received. 







Printed and bound by the 

University Herald Press,, 
Ada, Ohio. 




DEDICATED 

TO MY INSTRUCTOR, 

Mrs. EVA MAGLOTT, A. M., 

PROFESSOR OF MATHEMATICS 

IN THE 

Ohio Normal University. 




J f 



PREFACE. 



An attempt is made in this small volume to bring the 
science of Arithmetic and Geometry directly to the compre- 
hension of the learner, and to accomplish this end it is nec- 
essary to give a complete solution to the problems. This vol- 
ume is the outgrowth of a long course of study and long ex- 
perience in teaching. The experience of the author teaches 
him that most books of this class now in use treat the sub- 
ject in too brief or too difficult a manner. Most of our solu- 
tion books treat on problems that are out of the reach of com- 
mon school teachers. It is the aim of the author to' give 
them the solution of problems that they can comprehend. 

More time is spent on the study of Arithmetic in our 
schools than on any other one branch of study; and still the 
results are no better. It has been our aim to give solutions 
to every peculiar problem that usually gives our teachers 
trouble, and if we have omitted any they will be given space 
in later editions of this work. 

The method employed for the solution of problems in 
Stocks and Bonds and Commission, I received from Prof. Ed. 
M. Mills while he was my instructor. In my opinion, it is the 
best method. Many hints on the subject of mensuration were 
received from him. I also acknowledge my indebtedness 
to Prof. G. B. M. Zerr and Prof. J. C. Gregg for aiding me 
in some difficult work in trigonometrical regida falsi, or Double 
Position. 

I have taken problems from the School Visitor, from the 
Teachers' Review > published by Stonebrook & Maurer, from 

(v) 



vi PREFACE. 

the American Mathematical Monthly, published by Professors 
B. F. Finkel and J. M. Colaw, by the permission of these emi- 
nent mathematicians. I submit this book to my fellow teach- 
ers. Any correction will be thankfully received, knowing 
that it is almost impossible to put out a book whose first edi- 
tion is free from error. 

Crawfis College, Ohio, August, i8q8. 



CONTENTS. 

CHAPTER I. 

Page 
Proportion Discussed 1 

CHAPTER II. 

Analysis 4 

Time Problems 19 

CHAPTER III. 

Longitude and Time, and Latitude and Time .... 23 

Notes to Remember 26 

CHAPTER IV. 
Percentage 28 

CHAPTER V. 
Trade Discount 36 

CHAPTER VI. 
Profit and Loss 40 

CHAPTER VII. 
Commission 63 

CHAPTER VIII. 
Stocks and Bonds 74 

CHAPTER IX. 
Insurance. 86 

CHAPTER X. 

Interest 89 

True Discount 96 

Bank Discount 100 

Some Interesting Problems 103 

CHAPTER XI. 

County Examination Tests and Other Problems . . . 107 

Boat Problems » 133 

(vii) 



viii CONTENTS. 

CHAPTER XII. 

Mensuration 135 

Triangle 136 

Equilateral Triangle 162 

Formulas Used in Solutions 171 

Scalene Triangle . . . . 172 

Scalene Triangles Whose Areas are Integral .... 176 

Circle . 176 

Ellipse 196 

Catenarian Curve . . . 197 

Cycloid 198 

Curve of Pursuit 200 

Parallelograms and Solids 202 

The Origin of tt 234 

Tetrahedron 238 

CHAPTER XIII. 

Problems 240 

CHAPTER XIV. 

Rules of Mensuration , 253 

Points for the Student 254 

Journals 255 



FAIRCHILD'S 

SOLUTION BOOK 



CHAPTER I. 



PROPORTION DISCUSSED. 



1. There are few subjects in Arithmetic more easily mas- 
tered and more valuable in developing the reasoning powers 
than Proportion, and at the same time few students are in love 
with this subject. Students should put more time on this 
fascinating part of Arithmetic. 

2. Ratio is the quotient arising from dividing one quantity 
by another of the same kind. 

Thus, the ratio of 8 apples to 2 apples is 8 apples -7- 2 
apples = 4. We may inquire how many times 2 apples are 
contained times in 8 apples, but not how many times 2 
peaches are contained in 8 apples, because no number of 
times one will produce the other, i. e., no one of these quanti- 
ties can be multiplied and the other obtained. .'. There is no 
ratio of 8 apples to 2 peaches. 

3. Proportion is the equality of two ratios. 

The ratio of 8 apples to 2 apples is 4, or a PP es = 4 
rr rr 2 apples 

The ratio of 12 peaches to 3 peaches is 4, or " peac , es = 4. 
r r 3 peaches 

-p, , , r -.- 8 apples 12 peaches 

Then, by definition, rt , — = -^- — : . 

' 2 apples 3 peaches 



2 FAIRCHIL&S SOLUTION BOOK. 

Or, 8 apples : 2 apples:: 12 peaches : 3 peaches. The first 
and fourth terms are the extremes. The second and third are 
the means. Now, in any problem in Simple Proportion, if 
three of these four quantities be given, the fourth can be 
found. It is evident that if we have the two means and one 
extreme, we can find the other extreme by dividing the prod- 
uct of the means by the given extreme. In the same manner 
we can find the other mean if we have the two extremes and 
one of the means. 

4. In solving problems by Simple Proportion, we must 
always have the following relations: 

Greater : Less :: Greater : Less 
or, Less : Greater :: Less : Greater 

Take for the third term the number which is of the same 
kind as the answer sought. That is, if peaches are required, 
put for the third term the given number of peaches. After 
this is done, we must ascertain whether more or less of the re- 
quired quantity is necessary to fulfill the conditions of the 
problem; if less, we have, greater : less :: greater : less; but 
if greater, less : greater :: less : greater. 

6. We shall illustrate the method. 

If 28 men niow a field of grass in 12 days, how many men will be 
required to mow it in 8 days? 

If 28 men mow a field in 12 days, will it take a greater or 
less number of men to mow it in 8 days? Answer, greater; 
therefore, we must have: 

Less : Greater :: Less : Greater 

8 days : 12 days :: 28 : x = 42 men. 

7. The author believes this to be the simplest method of 
illustrating Compound Proportion. 

If 28 men dig- a ditch 120 rods long, 15 feet wide and 12 feet deep, 
how many men will dig a ditch 360 rods long, 9 feet wide and 10 feet 
deep? 

If 28 men dig a ditch 120 rods long, will it take a greater or 
less number of men to dig a ditch 360 rods long? Answer, 
greater. Then we have: 



PROPORTION DISCUSSED. 3 

Less : Greater :: Less : Greater 
120 rods : 360 rods :: 2fmen : x 

If 28 men dig a ditch 15 feet wide, will it take a greater or 
less number of men to dig a ditch 9 feet wide? Answer, less. 

Greater : Less :: Greater : Less 
15 feet : 9 feet:: 28 men : x 

If 28 men dig a ditch 12 feet deep, will it take a greater or 
less number of men to dig a ditch 10 feet deep? Answer, less. 

Greater : Less "Greater : Less 
12 feet : 10 feet :: 28 men : x 

Now write out the three simple proportions in the form of 
a compound proportion. Thus: 

120 rods : 360 rods ) 

15 feet : 9 feet [ :: 28 men : x = 42 men. 
12 feet : 10 feet ) 

A few lessons given on the above method will fix Propor- 
tion firmly in the student's mind. 



CHAPTER II. 



ANALYSIS. 



8. Arithmetical Analysis is the process of developing prob- 
lems by comparison of them through their relations to the 
unit. 

NoTK. — Nothing- is of more importance than Analysis. The reason 
why Arithmetic is so difficult to many students is that they do not pay 
proper attention to arithmetical analysis. 

9. f- -r- f . Demonstrate. 

One fourth is contained in one unit 4 times, but 3 fourths 
are 3 times as much. Then it is plain that it is contained in 
a unit i of 4 times, or f of 1 time, and in f of a unit it is 
contained f of f, or f of 1 time. 

Rule. — Invert the divisor and multiply the fractions together. 

PROBLEM 1. 

What number is that, to which if you add f of itself the sum will 
be 20? 

Solution. 

(1) H = the required number. 

(2) J + f =^. 

(3) -V°- = 20. 

(4) \ = A of 20 = 2. 

(5) \ — 7 times 2 = 14. 
.-. 14 = required number. 

PROBLEM 2. 
If f of a yard cost $f , what will f of a yard cost? 

Solution. 

(1) If f of a yard cost $f , £ of a yard will cost \ of f , or $\. 

(2) |, or a yard, will cost 3 times \ = $f . 



ANALYSIS. 



(3) Hence, £ of a yard will cost |- of ff 
.'. -§- of a yard cost 



PROBLEM 3. 

A and B tog-ether have $930; f of A's money equals f of B's: how 
much has each? 

Solution. 

(1) f A's = | B's. 

(2) i A's = i of | B's = A B's. 

(3) | A's = 4 times ^B's = |f B's, value of A's in terms 

of B's. 

(4) ff = B's money. 

(5) -ff = A's money. 

(6) if-+-M = U = what both have. 

(7) $930 = what both have. 

(8) U = $930. 

(9) tV = 3 l r of$930 = $30. 

(10) |f = 15 times $30 = $450. 

(11) f| = 16 times $30 = $480. 

.-. $450 = A's money; $480 = B's money. 

PROBLEM 4. 

A man owning ^ of a farm sold | of his share for $1810: what was 
the farm worth at that rate? 

Solution. 

(1) | of j\ = T ^ = part sold. 

(2) $1840 = value of part sold. 

(3) A = $1840. 

(4) T V = | of $1840 =.$920. 

(5) ff = 15 times $920 = $13800. 

.-. The farm is worth $13800. 

PROBLEM 5. 

$20 is f of the cost of a barrel of whisky: what did it cost? 

Solution. 

(1) f = the cost of the whisky per barrel. 

(2) | of the cost = $20. 

(3) I of the cost = | of $20 = $10. 

(4) | of the cost == 5 times $10 = 

,*. $50 = cost of whisky. 



6 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 6. 

I spent f of my money and then received $36; after losing- § of 
what I then had, I had left $2 less than I had at first: what sum had I 
at first? 

Solution. 

(1) -i-l = my money. 

(2) f of ^-f my money = f of my money. 

(3) -ff my money — f of my money = J of my money. 

(4) ^ + $36 = what I had by second condition. 

(5) f of (i of my money + $36) = -ft of my money + $27 

= what 1 lost. 

(6) (i of my money + $36) — (-ft of my money + $27) = 

-ft of my money + $9. This is what I had left, but 
this is $2 less than what I had at first. 

(7) .*. if my money = (ft of my money + $9) + $2. 

(8) Hence, ft = $11. 

(9) ft = ft of $11 = $1. 

(10) i| my money = 12 times $1 = $12. 

.*. $12 = the money I had at first. 

PROBLEM 7. 

Nine men can do a piece of work in 8|- days: how many days may 
3 men remain away and yet finish the work in the same time by bring- 
ing 5 more with them? 

Solution. 

(1) 8J days = time in which 9 men can do the work. 

(2) 8-J days X 9 = 75, the number of days work. 

(3) 8J days X 6 = 50, the number of days work that the 6 

men do who work all the time. 

(4) 75 days — 50 days = 25 days. 

(5) Hence, the 8 men do 25 days work in 25 days -f- 8 = 3J 

days. 

(6) 8£ days — 3J days = 5& days. 
.*. They can stay away 5 5 6 ¥ days. 

PROBLEM 8. 
If | of 6 be 3, what will } of 20 be? 

Solution. 

(1) 1 of 6 = 4. 

(2) £ of 20 = 10. By the condition of the problem, 

(3) 4 = 3. 

(4) 1 - I 



ANALYSIS. 7 

(5) 10 = 10 times f, or 7J. 

PROBLEM 9. 

John, William and Frank can do a piece of work in 30 days. If 
John does £ as much as William and William § as much as Frank, how 
long- will it take each to do it? 

Solution. 

(1) f = account of work that Frank does. 

(2) I of | Frank's work = -§• of William's work. 

(3) | of | William's work = \ John's work. 

(4) Hence, all do § + | + \ = V;\ or 2\ times Frank's work. 

(5) In 1 day Frank does x % of .^ = B * ? of the work. 

(6) .it Frank can do in J?-| -J- ^, or 65 days. 

(7) j; William's work = 65 days. 

(8) I = \ of 65 = s. 

(9) | = 3 9, I days. 

(10) ^ John's work = 2 times 65 days = 130 days. 

.". John can do the work in 130 days; William in 97-^ days; 
Frank in 65 days. 

NOTE. — We can readily see that Frank does 2 times as much as 
John and I times as much as William. 

PROBLEM 10. 

A pole, the leng-th of which is 39 feet, is in the air and water; | of 
the length in the air + 6 feet equals li times the leug-th in the water: 
find the length of each part. 

Solution. 

(1) | = length of the part in the air. 

(2) | of \ the length of the pole in the air = £ of the length 

of the pole in the air. 

(3) £ + 6 feet = U, or f the length in the air + 4 feet = the 

length in the water. 

(4) /. | + f + 4 = length of the pole. 

(5) | + | + 4 = .39 feet. 

(6) I = 35 feet. 

(7) \ — i of 35 feet = 5 feet. 

(8) i = 4 times 5 feet = 20 feet. 

(9) 39 feet — 20 feet = 19 feet. 

.*. 20 feet is the length in the air, and 19 feet the length 
in the water. 



8 FAIRCHIL&S SOLUTION BOOK. 

PROBLEM 11. 

A and B can do a piece of work in 20 days, A and C in 15 days, B 
and C in 12 days. In what time can all do it? Each? 

Solution. 

(1) 20 days = time it takes A and B to do the work. 

(2) -^o = part they do in 1 day. 

(3) 15 days = time it takes A and C to do the work. 

(4) y 1 ^ •=. part they do in 1 day. 

(5) 12 days = time it takes B and C to do the work. 

(6) ^ == part they do in 1 day. 

(7) Then, ■^ r + : A + A = i=? P art A and B > A and C, and 

B and C do in 1 day = twice the work A, B and C do 
in 1 day. 

(8) y\ = ^ of -J- == part A, B and C do in 1 day. 

(9) {$— part A, B and C do in fg- -7- T \, or 10 days. 

(10) ^ — -Jq- =^o — part A, B and C do in 1 day — part A 

and B do in 1 day = part C does in 1 day. 

(11) -I® == part C does in -f-g- -7- ^, or 20 days. 

(12) yV — yV = ^0 =: P art A, B and C do in 1 day — part A 

and C do in 1 day = part B does in 1 day. 

(13) fg- = part A does in %% -7- -^ = 30 days. 

(14) yV — iV — A — P art A, B and C do in 1 day — part B 

and C do in 1 day = part A does in 1 day. 

(15) f-g- = part A does in f-g- -7- ^5-, or 60 days. 

.". A, B and C do the work in 10 days, A in 60 days, B in 
30 days, and C in 20 days. 

PROBLEM 12. 

A man and a boy can do a certain work in 10 hours; if the boy 
rests 2f hours, it takes them 11^ hours: in what time can each do the 
work? 

Solution. 

(1) Hi hr. — 2| hr. = 8£f hr. = time they both work to- 

gether by second condition of the problem. 

(2) 10 hr. = time in which they both do the work. 

(3) Y^ = part they do in 1 hour. 

(4) 8-H -7- 10 — |fj — p ar t they do in 8££ hours. 

(5) -ftt — 1 H = -$£$ = P art tne man does while the boy 

rests in 2-J hours. 

(6) t^j- -7- 2-f == T !g, part the man does in 1 hour. 

(7) yf = part the man can do in yf -7- 18, or 18 hours. 

(8) tV — A = t|-o» P art the D °y does in 1 hour. 



ANAL YSIS. 9 

(9) i-f-g- = part the boy does in -ff-g- 4- T fo> or 22-J hours. 
.'. It will take the man 18 hours and the boy 22^- hours. 

PROBLEM 13. 

A can do as much work in 2 days as B can in 2-£; and B can do as 
much in 2 days as C can do in 2£. They work tog-ether and earn $91.50: 
what amount do they each receive? 
A Solution. 

(1) Let 20 units or parts represent A's work in 2 days, 

(2) In 1 day A can do \ of the same work, of which B can do 

f in a day. 

(3) .". \ of 20 units = 10 units, part that A does in 1 day. 

(4) | of 20 units = 8 units, part that B does in 1 day. 

(5) In 1 day B can do \ of the same work, of which C can do 

f in a day. 

(6) 16 units = part that B can do in 2 days. 

(7) I of 16 = 6f , part that C can do in a day. 

(8) .'. The parts are 25, 20 and 16; and A receives £§■ of $91.50, 

or $37.50; B, |? of $91.50, or $39; and C, \\ of $91.50, 
or $24. 

.-. A receives $37.50; B, $30; C, $24. 

PROBLEM 14. 

If f of A's turkeys, plus \ of B's, equals 900, how many turkeys has 
each, provided £ of B's number is twice f of A's number? 

Solution. 

(1) f = A's number of turkeys. 

(2) | of f = | X 2 = f, twice A's number of turkeys = f of 

B's number of turkeys. 

(3) | of A's + | or f of B's = f = 900 turkeys. 

(4) \ — -I- of 900 = 150 turkeys. 

(5) f == 3 times 150 turkeys = 450, A's turkeys. 

(6) | of 450, A's number = 300 turkeys. 

(7) j of B's turkeys = 300 X 2 = 600 turkeys. 

(8) i = I of 600 = 200 turkeys. 

(9) i = 4 times 200 = 800 turkeys. 

/. A has 450 turkeys; B, 800. 

PROBLEM 15. 
A man after doing- f of a piece of work in 30 days eng-ag-es an 
an assistant and both tog-ether complete it in 6 days: in what time 
could the assistant do it alone? 



10 FAIR CHILD'S SOLUTION BOOK. 

Solution. 

(1) f = the work., 

(2) f = the part the man does in 30 days. 

(3) f = 30 days. 

(4) i = i of 30 days = 10 days. 

(5) f = 5 times 10, or 50 days, the time in which the man 

can do the whole work. 

(6) f — f = f, the part the man and assistant do in 6 days. 

(7) f = 6 days. 

(8) i = i of 6, or 3 days. 

(9) f — 3 times 3, or 15 days, the time in which the man and 

assistant can do the whole work. 

(10) 5^ = the part the man does in 1 day. 

(11) ^ = the part the man and assistant do in 1 day. 

(12) ff# = what the assistant does in f|§ -l- ^ — 21f days. 

.'. The assistant can do the work alone in 21f days. 

PROBLEM 16. 

Six men can do a piece of work in 4^ days; after working- two days, 
how many men must join them to complete it in 3| days from the time 
of beginning the work? 

Solution. 

(1) 4^ days = time in which 6 men can do the work. 

(2) 4^ days X 6 = 26 days, the time 1 man can do it. 

(3) 2 days X 6 = 12 days, time they all work from the start. 

(4) .*. 26 days — 12 days, or 14 days work remain to be done 

in 3f days — 2 days = If days. 

(5) 14 days -4- If = 10, the number of men required to finish 

the work. 

(6) Hence, 10 men — 6 men = 4 men. 

.*. 4 men must join them. 

PROBLEM 17. 

If 3 times Susan's ag-e is f of Daisy's age, in how many years will 
Daisy be just twice as old as Susan? {Raub's Arith., p. 323.) 

Solution. 

(1) Let D = Daisy's present age, and S = Susan's. 

(2) By the condition of the problem, D = 8S. 

(3) .'. 2S + tne required number of years = D, or 8S. 

(4) Hence, the required number of years must be 6S, or 6 

times Susan's present age. 



ANALYSIS. 11 

(5) It is no matter what Susan's present age may be, when it 

is increased by 6 times itself it will be \ of Daisy's. 

(6) Suppose Susan's age is 2, Daisy's 16. 

(7) Susan's age added to 6 times itself = 14 years; this is \ 

of Daisy's, the latter being 16 + 12, or 28 years. 

Note. — Many teachers stumble over this problem. 

PROBLEM 18. 

The age of Jane is twice the age of Mar3% and % of Mary's age plus 
44 years equal 2 J times the age of Jane: what is the age of each? 

Solution. 

(1) | = Mary's age. 

(2) J5 - = Jane's age. 

(3) f of f Mary's age =| of Mary's age. 

(4) -§ of Mary's age -f- 44 years = 2? f times the age of Jane. 

(2*x v = W 

(5) .-.f+44= V-. 

(6) Hence, V = 44 years. 

(7) \ = A of 44 = 2 years. 

(8) .'. g Mary's age = 5 times 2 years, or 10 years. 

(9) 10 years X 2 = 20 years, the age of Jane. 
.". Mary is 10 years old and Jane 20. 

PROBLEM 19. 

C's age at A's birth was 5| times B's, and now is the sum of A's 
and B's, but if A were now 3 years younger and B 4 years older, A's 
age would be % of B's: hud their ages. 

Solution. 

(1) Since C's age at A's birth -j- A's present age = A's pres- 

ent age -f~ B's; 

(2) .'. C's age at A's birth = B's present age. B) r the second 

condition, A's age -f- 3 = -j (B's + 4), from which A's 
age = j B's age -f- 6 years. 

(3) Now, the difference between A's and B's present age = 

B's at the birth of A. 

(4) .'. \ B's present age — 6 years = B's age at A's birth, 

and 5^ (\ B's age — 6 years) = y- B's present age, 
from which \ l ~ B's age — 33 years = B's age, or 88 
years. 

(5) .'. C's age was 88 years at A's birth, B's 88 -r- 5| = 16 

years; and A's present age is 88 — 16 = 72 years, B's 
88, and C's 88 + 72 = 160 years. 

.'. A is 72 years old, B is 88, and C 160. 



12 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 20. 

A man engaging- in trade lost f of his money invested, after which 
he gained $840, and then had $3680: how much did he lose? 

Solution. 

(1) f = what he had at first. 

(2) f = part lost. 

(3) * — * = *, part left. 

(4) $840 = amount gained afterward. 

(5) | + $840 = amount he had after gaining $840. 

(6) $3680 = amount he had after gaining $840. 

(7) f + $840 = $3680. 

(8) f = $2840. 

(9) i= J of $2840 = $946| . 

(10) £•=" 5 times $946| = $4733J, his money. 

(11) -J of $4733^ = $18931. 
.-. He lost $1893^. 

PROBLEM 21. 

John spent f of f more than half his money, and had $210 left: how 
much had he at first? 

Solution. 

(1) | of i = A- 

(2) -§ = his money. 

(3) \ = half of his money. 

(4) *of* = A. 

(5) i + A = It. P^t spent. 

(6) »—■.» = A. P art left - 

(7) A = $210. 

(8) 3V = \ of $210 = 

(9) ffl = 30 times $30 
= what he had at first 



NoTK. — 1\ more than \ is -fa of \ plus \. 

PROBLEM 22. 
If f of the time past noon, plus 3^ hours, equals f of the time to 
midnight, plus 2J- hours, what is the time? 

Solution. 

(1) \ = time past noon. 

(2) 12 hours = time from noon to midnight. 

(3) f of f time past noon = f of time past noon. 



ANAL YSIS. 13 

(4) 12 hours — \ time past noon = time to midnight. 

(5) f of time past noon -f~ 3-J- hours == -J of (12 hours — \ 

time past noon) + 2 J- hours. 

(6) .*. f of time past noon -f- 3| hours = 8 hours — A of 

time past noon + 2-J- hours. 

(7) -}-J of time past noon = ^- hours. 

(8) iV = tt of ¥■- or tot h ours. 

(9) \\ = 12 times -fife - 4^4 hours. 

.*. 4j-| hours, or 4 hr. 49 min. 24j-| sec. is the time past 
noon. 

PROBLEM 23. 

A man spent \ more than § of his money and had $500 left: how 
much did he have? 









Solution. 




(1) 


i = 


his money. 




(2) l of 


1 = *. 






(3) 


i + 


A = If. or A» the amount 


spent 


(4) 


1 _ 
1 0" 


-A = A. 


the amount left. 




(5) 


l - 

10" 


= $500. 






(6) 


10 - 

10 - 


= 10 times 


$500 = $5000. 




•. He 


had 


$5000. 







PROBLEM 24. 

Ten years ago the age of John was £ of the age of Henry, and ten 
years hence the age of John will be f of the age of Henry: find the 
age of each. 

Solution. 

(1) I = Henry's age 10 years ago. 

(2) I = John's age 10 years ago. 

(3) I + 20 years = Henry's age 10 years hence. 

(4) f + 20 years = John's age 10 years hence. 

(5) f of (| + 20 years) = f Henry's + 16$ years. 

(6) |- Henry's + 16| years = J Henry's + 20 years. 

(7) -fe Henry's = 3J years. 

(8) }-| = 12 times 3£ years = 40, Henry's age 10 years ago. 

(9) I of 40 years = 30 years, John's age 10 years ago. 

(10) 40 years + 10 years = 50 years, Henry's present age. 

(11) 30 years + 10 years = 40 years, John's present age. 

.*. Henry's age is 50 years, John's age 40 years. 



14 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 25. 

A hare starts 30 yards before a hound, but is not seen by him till 
she has been running- 40 seconds. The hare runs at the rate of 8 miles 
an hour, and the hound pursues her at the rate of 10 miles per hour: 
how long- will the chase continue, and how far must the hound run? 

Solution. 

(1) 10 miles = distance the hound runs in 1 hour. 

(2) 8 miles = distance the hare runs in 1 hour. 

(3) 10 miles — 8 miles = 2 miles, distance the hound gains 

on the hare in 1 hour. 

(4) 1760 yards = the number of yards in 1 mile. 

(5) (1760 yards X 2) -f- (60 X 60) = f f-J yards, distance the 

hound gains on the hare in one second, 

(6) 30 yards + (40 X 8 X 1760) 4- (60 X 60) = 186f yards, 

the distance to be gained by the hound. 

(7) 186| yards H- ff | = 190|| seconds, or 3 min. 10-|| sec. 

.*. 186| yards is the distance the hound runs, and 3 min-. 
10-|-| sec. is the time of the chase. 

PROBLEM 26. 

A fox is 120 leaps before a hound and takes 5 leaps to the hound's 
2, but 4 of the hound's leaps equal 12 of the fox's: how many : leaps 
must the hound take to catch the fox? 

Solution. 

(1) 2 of the hound's leaps = 6 of the fox's, but while the 
hound makes 6 of the fox's the fox takes only 5, hence, 
the hound gains 1 of the fox's leaps in making 2, and 
in 120 the hound must take 120 X 2 = 240 leaps. 

.'. The hound takes 240 leaps to catch the fox. 

PROBLEM 27. 

A fish's head is 4 inches long-; its tail is as long- as its head and \ 
of its bod} r ; the body is as long as its head and tail: what is its length? 

Solution. 

(1) f = length of body. 

(2) 4 inches == length of head. 

(3) 4 inches + \ of body = tail. 

(4) 8 inches + \ of body = body. 

(5) \ of body = 8 inches. 

(6) \ of body = \ of 8 in. = 4 in. 

(7) f length of body = 3 times 4 in. = 12 in. 

(8) 4 in. + \ of body (or tail) = 8 in. 



ANALYSIS. 15 

(9) .-. 4 in. + 8 in. + 12 in. = 24 in., length of fish. 
.*. The fish is 24 inches long. 

PROBLEM 28. 

There is coal now on the dock and coal is running- on also from a 
chute at a uniform rate. Six men can clear the doek in one hour, but 
11 men can clear it in 20 minutes: how long- will it take 4 men? 

Solution. 

(1) Suppose 1 man's work to be 6 tons. 

(2) Then 6 tons X 6 = 36 tons, whole quantity on dock 1 hr. 

(3) 11 X 6 X \ hr. (or 20 minutes) = 22 tons on dock in 20 

minutes. 

(4) 36 tons — 22 tons = 14 tons, quantity running on in 1 — 

^ = | of an hour. 

(5) 14 tons —- I = 21 tons run on in 1 hour. 

(6) 36 tons — 21 tons = 15 tons, original quantity on dock. 

(7) 21 tons -r- 6 = 3V, or 3* times a man's work to clear what 

is running on. 

(8) 4 — Z\ = \ of 1 man's work left to clear original quan- 

tity, or 15 tons. 

(9) Half a man's work being 3 tons an hour, it will take 15 -r 

3, or 5 hours. 

.'. It will take 4 men 5 hours to clear the dock. 

PROBLEM 2!). 

A man at his marriage agreed that if at his death he should leave 
only a daughter, his wife should have \ of his estate; and if he should 
leave a son, she should have }. He left a sou and a daughter. What 
fractional part of the estate should each receive, and how much was 
each one's portion, the estate being worth $4000? 

Solution. 

(1) The wife in the first condition was to have |, or 3 times 

as much as the daughter. 

(2) In the second condition the son was to have |, or 3 times 

as much as the wife. 

(3) Daughter has 1 part, wife 3 parts, son 9 parts, or ^ s -f- ^ 

-|- ^ — ]:] = whole estate. 

(4) fa of $4000 = $307^, daughter's share. 

(5) VV of $4000 = $923^, wife's share. 

(6) ^ of $4000 = $2769^, son's share. 

.-. Daughter's share is $307 /V, wife's $923^, son's $2769*3. 



16 FAIR CHILD'S SOL UTION B O OK. 

PROBLEM 30. 

A was engaged for a year at $80 and a suit of clothes; he served 
7 months and received for his wages the clothes and $35: what was 
the value of the clothes? 

Solution. 

(1) -Jf = the value of the clothes. 

(2) i| + $80 = wages for 12 months. 

(3) T V of (if + $80) = t l + $6f , wages for 1 month. 

( 4 ) (tt + $ 6 t) X 7 = T V +'$46f, wages for 7 months. 

(5) if +'$35 = wa g es f° r 7 months. 

(6) if + $35 = A + $46*. 

(7) A = »1*. 

(8) tSr = 5 of $ n f = 

(9) if = 12 times $2^ 

.*. $28 was the value of the clothes. 



PROBLEM 31. 

A teacher has books in three rooms; f of the number in the first 
room equals f of the number in the second room, and f of the number 
in the second room equals f of the number in the third room. If the 
entire number is 651, how many are in each room? 

Solution. 

(1) $ of the number in the first room = f the number in the 

second room. 

(2) ^ of the number in the first room == f of f \== -§ of num- 

ber in second room. 

(3) f, or number in first room = 3 times | = f of number 

in second room. 

(4) Then ■§■ = number in second room. 

(5) f of number in 3d room = f of number in 2d room. 

(6) \ of number in 3d room — J of -J = -§- of no. in 2d room. 

(7) J, or number in 3d room = 4 times f = f of number in 

2d room. 

(8) .-. f + | + -J = -W- = 651 » number in the 3 rooms. 

(9) ^ = drr of 651 = 3. 

(10) ff = 81 times 3, or 243, number in 1st room. 

(11) ^f = 72 times 3, or 216, number in 2d room. 

(12) -ff = 64 times 3, or 192, number in 3d room. 

.*. There are 243 books in 1st room, 216 in 2d, 192 in 3d. 



ANALYSIS. 17 

PROBLEM 32. 

Four masons, A, B, C and D, engage to build a wall; A, B and C 

can build it in 18 days; B, C and D in 20 days; A, C and D in 24 days; 

and A, B and D in 27 days: in what time can they build it jointly and 

separately? {Schuyler's H. A., p. 421, prod. 18.) 

Solution. 

(1) 18 days = time it takes A, B and C to build the wall. 

(2) .*. y 1 ^ = part they do in 1 day. 

(3) 20 days = time it takes B, C and D to build the wall. 

(4) -Jq. = part they do in 1 day. 

(5) 24 days = time it takes A, C and D to build the wall. 

(6) -^ — part they do in 1 day. 

(7) 27 days = time it takes A, B and D to build the wall. 

(8) 2V — P art they do in 1 day. 

(9) .*. iV + -2 J o + 2V + -if = -iWo* Part A, B and C; B, C and 

D; A, C and D; and A, B and D do in 1 day = 3 times 
the work A, B, C and D do in 1 day. 

(10) .-. i- of jVsV = 3^ , part A, B, C and D do in 1 day. 

(11) ^ 2 Vo — A = 32b. P art A, B, C and D do in 1 day - 

part A, B and C do in 1 day = part D does in 1 day. 

(12) f|fo = part D does in f|M ^ *ih = I™!? days. 

(13) uWo — A = rf io» Part A, B, C and D do in 1 day — 

part B, C and D do in 1 day = part A does in 1 day. 

(14) mi = Part A does in ffffl -r ^| = 87JI days. 

(15) sVA — A = M o» Part A, B, C and D do in 1 day — 

part A, C and D do in 1 day = part B does in 1 day. 

(16) MU = Part B does in ffi# + rff- tf = 50f days. 

(17) AVo — A = 3ifo> Part A, B, C and D do in 1 day — 

part A, B and D do in 1 day = part C does in 1 day. 

(18) mi = Part C does in fit* + A<ft = *V* days. 

(19) «n = Part A, B, C and D do in |||| -r 3 Wo = l&AV 

days. 

.*. They can jointly do the work in 16/^ days. A can do 
the work in 87$-£ days, B in 50| days, C in 41-^ days, 
and D in 170jf days. 

PROBLEM 33. 

A, B and C dine on 8 loaves of bread; A furnishes 5 loaves; B, 3 
loaves; C pays the others 8 dimes for his share: how must A and B 
divide the money? 

Solution. 

(1) 8 = the number of loaves they eat. 

(2) 8 -J- 3 = 2f , the number each eats. 



18 FAIRCHILD'S SOLUTION BOOK. 

(3) 5 loaves — 2f == 2J- loaves, what A furnished for C. 

(4) 3 loaves — 2-| loaves = \ loaf, what B furnished for C. 

(5) $ : h or A ' s portion is to B's as 7 : 1. 

.-. A's = i of 8 d. = 7 d., A's portion of the 8 d.; and B's 
is -J- of 8 d. = 1 d., B's portion. 

PROBLEM 34. 

A is twelve years old, and if to his ag-e be added f of the ag-es of A 
and B, the sum will be B's ag-e: what is B's ag-e? 

Solution. 

(1) Let f = B's age. 

(2) 12 + f of (12 + f, B's age) = f, B's. 

(3) 12 + Y + f B's = f, B's. 

(4) 96 + 60 = 3 B's, or 3 B's = 156. 

(5) B's age = 52 years. 

.*. 52 years = B's age. 

PROBLEM 35. 

What is the time of day when f of the time past noon equals || of 
the time till midnight? 

Solution. 

(1) f = time past noon. 

(2) 12 hours = time from noon to midnight. 

(3) | of -§, time past noon = ^f of (12 — f, time past noon.) 

(4) -§■ of time past noon = 1 6 8 2 r i rL - ijL of time past noon. 

(5) 10 times past noon = 168 hours — 14 times past noon. 

(6) 24 times past noon = 168 hours. 

.*. Time past noon = 7 hrs. 

PROBLEM 36. 

A lady spent $40 more than ^ the money in her purse; then $30 
more than £ of the remainder; then $10 more than ^ the remainder, 
after which she had $32: how much had she at first? 

Solution. 

(1) Let f == her money at first. 

(2) \ of \ of her money = \ of her money. 
. (3) \ + $40 = amount spent. 

(4) -§- her money — (| her money + $40) = \ her money — 

$40, the remainder. 

(5) \ of \ of her money — $40 = \ of her money — $y + 

$30 = \. of her money + $- 5 3 -, amount spent. 



ANALYSIS. 19 

(6) \ of her money + $40 — \ or her money + $4f = \ of 

her money — $-HP"> remainder. 

(7) \ of £ of her money — $^- + $10 = T V of her money — 

$f|, amount spent. 

(8) J of her money — $- 1 T %°- — ^ of her money — $f| = J of 

her money — $ 6 r2 0_ , remainder. 

(9) \ of her money — $- ( y 3 2 - = $32, or \ of her money = $84|. 
(10) | her money = 4 times $84| = $338. 

.-. She had $338 at first. 



PROBLEM 37. 

Charles, when asked his ag-e, replied: "My father was born in 1843 
and my mother in 1847. The sum of their ages at the time of my birth 
was 5 times my age in 1887." In what year will Charles be 25 years of 
age? 

Solution. 

If the father was born in 1843 and the mother in 1847, the 
sum of their ages in 1887 was 84 years; and since the sum of 
their ages at Charles's birth was 5 times his age in 1887, and 
the parents each increased in age as fast as he did, in 1887 the 
sum of their ages must have been 5 + 1 + 1, or 7 times the 
age of Charles; hence, he was 84 -r 7, or 12 years old in 1887, 
making the time of his birth 1875, and in 1900 he will be 25, if 
living. 



TIME PROBLEMS. 



PROBLEM 38 

At what time between 5 and 6 o'clock will the hour and minute 
hands be together? 

Solution. 

(1) Let | = distance the hour hand moves. 

(2) 2 ./ = the distance the minute hand moves. 

(3) \ 4 — f = A/ distance gained by minute hand. 

(4) 25 minutes = distance gained by minute hand. 

(5) \ 2 = 25 minutes. 

(6) \ = ^ of 25 minutes = |f minutes. 

(7) 2£ — 24 times ff = - 6 ^ min. = 27-/ f min. 
.'. 27^1 minutes past 5 = time required. 



20 FAIRCHILD'S SOLUTION BOOK, 

PROBLEM 39. 

At what time between 7 and 8 will the hour and minute hand of a 
clock be opposite each other? 

Solution. 

(1) Let f == distance moved by the hour hand. 

(2) \ 4 - = distance moved by the minute hand. 

(3) %£. — f = \ 2 - = distance gained by minute hand. 

(4) \ 2 - = 5 minutes. 

(5) \ = -^ of 5 minutes = -£% minutes. 

(6) V" = 24 times A = W min - = 5 tt min - 
.'. 5 T 5 T minutes past 7 = time required. 

PROBLEM 40. 

At what time between 6 and 7 o'clock is the minute hand the same 
distance from 9 as the hour hand is from 4? 

Solution. 

(1) f = distance moved by the hour hand. 

(2) \ 4 - == distance moved by the minute hand. 

(3) \ 6 = distance moved by both. 

(4) 35 minutes = distance moved by both. 

(5) V — 35 minutes. 

(6) \ — -^ F of 35 minutes = -ff minutes. 

(7) \ 4 = 24 times ff min. = -%$- min. = 32 T % min. 

.*. 32 T 4 3 minutes past 6 = time. 

Note. — Locate the minute hand at 12 and the hour hand at 6. When 
situated this way, the hour hand is 10 minutes from 4. If the hour hand 
remains stationary the minute hand must move 35 minutes to be within 
10 minutes of 9. The hour hand does not remain stationary, but moves 
farther from 4; hence the minute hand must stop short of 35 minutes 
just the same distance that the hour hand moves farther from 4. 

PROBLEM 41. 

At what time between 5 and 6 o'clock will the minute hand be at 
right angles with the hour hand? 

Solution. 

(1) -f = distance hour hand moves past 5. 

(2) \ 4 - = distance minute hand moves past 12. 

(3) \ 4 — f == ^ distance gained by minute hand. 

(4) -% 2 - = 10 min., or 40 min. 

(5) I = -fa of 10 min., or -^ of 40 min. = -|~| min., or 4$ min. 



ANALYSIS. 21 

(6) \ 4 = 24 times |f, or 24 times -f-J min. = 10-}-^- min., or 
43 T 7 T min. 
.'. The minute hand will be at right angles with the hour 
hand at 10^-J- minutes, or 43 T 7 r minutes past 5 o'clock. 

PROBLEM 42. 

At what time between 4 and 5 will the minute hand be £ of the dis- 
tance from 12 to the hour hand? 

Solution. 

(1) -f = distance moved by hour hand. 

(2) 20 minutes + £ = distance from 12 to hour hand. 

(3) £ of (20 min. + -f) = distance minute hand moves. 

(4) *£- == distance minute hand moves. 

(5) \ 8 = 5 minutes + -\. 

(6) \ 7 - = 5 minutes. 

(7) \ = ¥ y of 5 minutes = ^V minutes. 

(8) 4 4 8 = 48 times ^ T minutes = 5^ 7 minutes. 

.'. o^y minutes past 4 = time required. 

PROBLEM 43. 

At what time between 4 and 5 will the minute hand be as far from 
12 on the left side of the dial plate as the hour hand is from 12 on the 
right side? 

Solution. 

(1) -§ = distance moved by the hour hand. 

(2) %£■ = distance moved by the minute hand. 

(3) -\ 6 = distance moved by both. 

(4) 40 minutes = distance moved by both. 

(5) y = 40 minutes. 

(6) \ = 2 \- of 40 minutes = 4 g- minutes. 

(7) 2 2 4 = 24 times 4 g minutes = 36|J minutes. 

.*. 36}f minutes past 4 = the time required. 

PROBLEM 44. 

At what time between 5 and 6 will the minute hand be 10 minutes 
ahead of the hour hand? 

Solution. 

(1) :| z= distance hour hand moves while minute hand is 

moving to be 10 minutes ahead. 

(2) 2 2 4 = distance moved by the minute hand while the hour 

2 

2" 



22 FAIRCHILD'S SOLUTION BOOK. 

(3) \ 4 - — I = distance the minute hand gains on the hour 

hand. 

(4) 35 minutes = distance gained by minute hand. 

(5) -\ 2 - = 35 minutes. 

(6) I = 2 X 2 of 35 min. = f f minutes. 

(7) \ 4 - = 24 times f f> = 38 T \ minutes. 

.*. 38^ minutes past 5 = the time required. 

PROBLEM 45. 

Solar time at Cincinnati is 22 min. 16 sec. faster than Central Stan- 
dard time: what is the longitude of Cincinnati? {Putnam Co.) 

Solution. 
22 min. 16 sec, the difference of time := 5° 34', difference of 
longitude. Cincinnati is east of the Central time meridian, or 
90th. .'. 90° — 5° 34' == 84° 26' west longitude, or the longi- 
tude of Cincinnati. 

PROBLEM 46. 

At what time between 7 and 8 o'clock are the hour and minute 
hands of a watch together? {Brooks 1 Arithmetic.) 

Solution. 

(1) Let | = distance moved by the hour hand. 

(2) - 2 2 4 = distance moved by the minute hand. 

(3) - 2 2 4 - — \ = - 2 2 ? s distance gained by the minute hand. 

(4) 35 minutes = distance gained by the minute hand. 

(5) -\ 2 - = 35 minutes, and \ = ^ of 35 minutes, or \\. 

(6) - 2 2 4 = 24 times \\ or 38^ min. past 7. 



CHAPTER III. 

LONGITUDE AND TIME, AND LATITUDE AND 

TIME. 



10. The Latitude of a place is its distance from the equa- 
tor, north or south. It is measured in degrees, minutes and 
seconds, and can not exceed a quadrant, or 90°. 

11. The Longitude of a place is its distance east or west 
from a given meridian. It is reckoned in degrees, minutes 
and seconds, and can not exceed a semi-circumference, or 

180°. 

Rule. — When the latitudes or longitudes are both east or west, 
subtract the less from the greater ; when one is east and the other 
zv est, take their sum. 

Note.— When we add two longitudes, if their sum is 180°, it must 
be subtracted from 360° for the difference of longitude. 

PROBLEM 47. 

The longitude of New Orleans is 90° W., and of Geneva 6° 9'29"E.: 
what is the difference of longitude? {Brooks' 1 Arithmetic.) 

Solution. 
90° + 6° 9' 5" = 96° 9' 5", or the difference of longitude. 

PROBLEM 48. 

The longitude of Portland is 70° 13' 34" W., and of Mobile 88° 1' 29" 
W.: what is the difference of longitude? {Brooks' 1 Arithmetic.) 

Solution. 

88° 1' 29" — 70° 13' 34" = 17° 47' 55", the difference of lon- 
gitude. 

12. The circumference of the earth contains 360°; hence, 
the sun appears to travel through 360° in 24 hours. 

In 1 hour it travels ^ of 360° = 15°. 

(23) 



24 FAIR CHILD'S SOLUTION BOOK. 

In 1 minute it travels fa of 15° = 15'. 
In 1 second it travels fa of 15' = 15". 

13. To find the difference of time of two places. 

Rule. — Divide the difference of longitude by 15 and mark the 
quotient hr. min. sec. instead of ° ' ". 

14. To find the difference of longitude of two places. 
Rule. — Multiply the difference of time by 15 and mark the prod- 
uct ° ' " instead of hr. min. sec. 

PROBLEM 49. 

The longitude of New York is 74° 3' W., and of New Orleans 90° W.: 
required the difference in time. {Brooks' 1 Arithmetic.) 

Solution. 

(1) 90° — 74° 3' = 15° 57', the difference of longitude. 

(2) 15° 57' -r- 15 = 1 hr. 3 min. 48 sec, the difference of time. 

PROBLEM 50. 

The longitude of Philadelphia is 75° 9' 5" W., and of Cincinnati 84° 
29' 31" W. : what is the time at Cincinnati when it is 10 A.M. at Phila- 
delphia? {Brooks' 1 Arithmetic.) 

Solution. 

(1) Cincinnati is 84° 29' 31" W.; Philadelphia, 75° 9' 5" W. 

The latter is 9° 20' 26" farther east. 

(2) 9° 20' 26" -M5 = 37 min. 21ft sec. 

(3) /. 10 hr. — 37 min 21ft sec. = 9 hr 22 min. 38^- sec. A. 

M., time at Cincinnati. 

PROBLEM 51. 

The longitude of Rome is 12° 27' E}., and San Francisco 122° 26' 15" 
W.: what time is it in the latter place when it is 4 P. M. in the for- 
mer? {Brooks' 1 Arithmetic.) 

Solution. 

(1) Rome 12° 27' E., San Francisco 122° 26' 15" W., difference 

of longitude is 134° 53' 15". 

(2) 134° 53' 15" ~ 15 = 8 hr. 59 min. 33 sec. 

(3) 4 P. M. = 16 hr. 

(4) .-. 16 hr. — 8 hr. 59 min. 33 sec. = 7 hr. 27 sec. A. M., the 

time at San Francisco. 

PROBLEM 52. 
The difference of time between Philadelphia and Cincinnati is 
about 37 min, 20 sec: what is the difference of longitude? 

{Brooks' 1 Arithmetic.) 



LONGITUDE AND TIME. 25 

Solution. 

The difference of time, 37 min. 20 sec. X 15 = 9° 20', differ- 
ence of longitude. 

PROBLEM 53. 

When it is 11 A. M. at a place 30° E. of Greenwich, it is 3 hr. 44 
min. 20 sec. a. m. at Buffalo: what is the longitude of Buffalo? 

Solution. 

(1) 11 hr. — 3 hr. 44 min. 20 sec. = 7 hr. 15 min. 40 sec, dif- 

ference of time. 

(2) 7 hr. 15 min. 40 sec. X 15 = 108° 55', difference of long. 

(3) /. The longitude of Buffalo is 108° 55' — 30° = 78° 55'. 

PROBLEM 54. 

The longitude of Honolulu is 157° 52' W., and that of Sydney 151° 
11' E. When it is 5 min. after 4 o'clock on Sunday morning at Hono- 
lulu, what is the time at Sydney? {Ray's Higher.) 

Solution. 

(1) The difference of longitude is 157° 52' + 151° 11'= 309° 3'. 

(2) The difference of time is 309° 3' ~ 15 = 20 hr. 36 min. 

52 sec. This added to 4 hr 5 min. gives 24 hr. 41 min. 
52 sec. past midnight, or 41 min. 52 sec. A. M. Monday. 

PROBLEM 55. 

Locate and name a noted city whose sun time is 6:40 A. M. when i t 
is noon at Greenwich. 

Solution. 

The difference of time is 5J hr., and the city must be 5£ X 
15, or 80° west of Greenwich. Pittsburg, Pa., has that longi- 
tude. 

PROBLEM 56. 

If we consider a degree of longitude in our latitude 53 miles, how 
far and in what direction had I gone on a parallel when I found my 
watch 2 hr. 40 min. too fast? 

Solution. 

(1) The difference of longitude is (2 hr. 40 min.) X 15 = 40°. 

(2) Since my time was too fast, I must have gone west 40 X 

53, or 2120 miles. 

PROBLEM 57. 

The distance from Boston to Chicago is about 840 miles, and a de- 
gree of longitude at Boston contains about 51 miles; when it is noon at 
Boston what is the time at Chicago? 



26 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) The distance, 840 miles -^- 51 = 16 T 8 T °, diff. of longitude. 

(2) 16^° -f 15 — 1 T \% hr. before noon. 

PROBLEM 58. 

Yesterday I was in longitude 86° 18' W. and set my watch; to-day 
the sun is on the meridian at 36 min. past 11 o'clock: what is my longi- 
tude? 

Solution. 

The difference of time is 24 min., which reduced to degrees 
of longitude is 24 X 15 -r- 60 = 6°; then my longitude is now 
6° less, or 80° 18' W. Beaver Falls, Pa., has that longitude, 
and I am in the longitude of South Bend, Ind. 

NOTES TO REMEMBER. 

15. If the inclination of the earth's axis were 30° instead 
of 23|°, the width of the North Temperate Zone would be 90° 
— (30° + 30°) =30°, in fact, the remaining zones would be 30° 
wide. 

16. Standard Railroad Time Explained. — Fifteen degrees of 
longitude equal one hour of time. Railroad men, to avoid 
trouble and accident, have adopted the plan of making all 
places on the earth have the time of certain meridians, each 
one being 15° apart. The following meridians are used: 

0° is called Universal time, west 15° West African, 30° Cen- 
tral Atlantic, 45° East Brazilian, 60° La Plata, 75° Eastern, 90° 
Central, 105° Mountain, 120° Pacific, 135° East Alaskan, 150° 
Central Alaskan, 165° West Alaskan, 180° Transitional; east 
165° is called New Caledonian, 150° East Australian, 135° 
Corean, 120° East Asian, 105° Siam, 90° East Hindostan, 75° 
West Hindostan, 60° Ural, 45° Caucasus, 30° Bosporus, 15° 
Scandinavian. 

Columbus, O., Iyat. 39° 57' N., Long-. 83° 3' W. 

Washington, D. C, L,at. 38° 39' N., Long. 77° 3' W. Exact, 77° 2' 
48" W. 

St. Helena is an island in the Atlantic 15° 55' 26" S. Lat., and 5° 
42' 30" W. Long. 

Chicago has a longer day than Constantinople, because it is farther 
north by 50'. 

17. A Geographical or Nautical Mile (or Knot) is about 2025 
yards. 



LONGITUDE AND TIME. 27 

18. A Tropical Year is the time elapsed from the moment 
the sun leaves a star until it reaches it again. It equals 365 
days, 6 hours, 9 minutes and 9 seconds. 

19. A Solar Year is the time elapsed from the moment the 
sun leaves the vernal equinox till it reaches it again. It 
equals 365 days, 5 hours, 48 minutes and 48 seconds. 

20. The Anomalistic Year is measured from the time the sun 
leaves the perihelion till it reaches it again. It equals 365 
days, 6 hours, 13 minutes and 45.6 seconds. 

21. A Lunar Year is 12 lunar months. It equals 354 days. 

22. The Equator is a line passing around the globe midway 
between the poles. 

Note. — Some teach that the word "line" used in the above defini- 
tion is an imaginary line, yet a careful and thorough investigation 
will show that it is not an imaginary line, but real. All geometrical 
lines are real, according to definition. Investigate this matter and 
see if there has not been some fallacy in the definition both of the 
equator and also of small circles. 



CHAPTER IV. 

PERCENTAGE. 



23. Percentage is the process of computation in which 100 
is the basis of comparison. 

24. Per Cent {per, by, centum, a hundred) means by or on the 
hundred. 

25. The quantities used in Percentage are the Base, the 
Rate, the Percentage and the Amount or Difference. 

26. The Base is the number on which the percentage is 
computed. 

27. The Rate is the number of hundredths of the base 
which is to be taken. 

28. The Percentage is the result obtained by taking a cer- 
tain per cent of the base. 

29. The Amount or Difference is the sum or difference of the 
base and percentage. They may both be embraced under the 
general term, Proceeds. 

CASE I. 

PROBLEM 59. 
What is 12% of 475? 

Solution. 

(1) 100% = 475. 

(2) 1% = ^ of 475 - 4.75. 

(3) 12% = 12 times 4.75 = 57. 

.-. 12% of 475 = 57. 

PROBLEM 60. 
What is f % of $800? 

Solution. 

(1) 100% = $800. 
(28) 



PERCENTAGE. 29 



(2) 1% = T * T of 

(3) f% = | times 8 = $5. 

.-. f% of $800 = $5. 

PROBLEM 61. 
What is 10% of 20% of $13.50? 

Solution. 

(1) 100% = $13.50. 

(2) 1% = -rta of $13.50 = $.1350. 

(3) 20% = 20 times $.1350 = $2.7. 

(4) 100% = $2.7. 

(5) 1% = -rb of $2.7 = $.027. 

(6) 10% = 10 times $.027 = $.27. 

.-. 10% of 20% of $13.50 = $.27. 

PROBLEM 62. 

A man contracts to supply dressed stone for a court house for 
$119449, if the rough stone costs him 16 c. a cu. ft.; but if he can get 
it for 15 c. a cu. ft., he will deduct 3% from his bill: how many cu. 
ft. would be needed, and what does he charge for dressing - a cu. ft.? 

Solution. 

(1) 100% = $119449. 

(2) 1% = T U of $119449 = $1194.49. 

(3) 3% = 3 times $1194.49 = $3583.47, the deduction. 

(4) $.16 — $.15 = $.01 the deduction per cubic foot. 

(5) $3583.47 -=- $.01 = 358347, cubic feet. 

(6) $119449 = cost of 358347 cubic feet. 

(7) $119449 -T- 358347 = $.33J, cost of dressing 1 cubic foot. 

(8) $.33£ — $.16 =$.17J, cost of dressing per cubic foot. 

.*. 358347 = number of cubic feet, $.17J = cost of dress- 
ing per cubic foot. 

PROBLEM 63. 

48% of brandy is alcohol: how much alcohol does a man swallow in 
40 years, if he drinks a gill of brandy 3 times a day? 

Solution. 

(1) 365J = number of days in a year. 

(2) 365J X 40 = 14610 days. 

(3) 14610 X 3 gills = 43830 gills. 

(4) 100% = 43830 gills. 



30 FAIR CHILD'S SOLUTION BOOK. 

(5) 1% = -^ of 43830 gills = 438.3 gills. 

(6) 48% = 48 times 438 8 gills = 21038.4 gills. 

.-. He swallows 21038.4 gills, or 657 gal. 1 qt. 1 pt. 2.4 gi. 

PROBLEM 64. 

A man has $1200; he gave 30% to a soti, 20% of the remainder to 

his daughter, and so divided the rest among four brothers that each 

after the first had $12 less than the preceding: how much did the last 

receive? {Ray's Higher, p. ipi, prod. 25.) 

Solution. 
: (1) 100% = $1200. 

(2) 1% = t l> of $1200 = $12. 

(3) 30% = 30 times $12 = $360, son's share. 

(4) $1200 — $360 = $840, remainder. 

(5) 100% = $840. 

(6) 1% =^U of $840 = $8.40. 

(7) 20% = 20 times $8.40 = $168, daughter's share. 

(8) $840 — $168 = $672, amount to be divided between the 

4 brothers. 

(9) 100% = 4th brother's share. 

(10) 100% + $12 = 3d brother's share. 

(11) 100% + $24 = 2d brother's share. 

(12) 100% + $36 = 1st brother's share. 

(13) 100% + (100% + $12) + (100% + $24) + (100%+$36) 

= 400% + $72, amount the brothers receive. 

(14) 400% +$72 = $672 

(15) 400% = $600. 

(16) 1% =T Lof $600 = $1.50. 

(17) 100% = 100 times $1.50 = $150. 

.". The fourth brother received $150. 

PROBLEM ,65. 

What number increased by 20% of 3.5, diminished by 12|% of 9.6, 
gives 3£? {Ray's Higher, p. 191, prod. 26.) 

Solution. 

(1) 100% = the number. 

(2) 100% = 3.5. 

(3) 1% = T i of 3.5 = .035. 

(4) 20% = 20 times .035 = .7. 

(5) 100% = 9.6. 

(6) 1% = ttTo of 9.6 = -096. 

(7) 12i% _ 121 times .096 = 1.2. 



PERCENTAGE. 31 

(8) .*. 100% + .7 — 1.2 = 3.5. 

(9) 100% = 4. 
.*. The number is 4. 

CASE II. 

PROBLEM 66. 

$14 is what % of $175? 

Solution. 

(1) $175 = 100%. 

(2) $1 =^ 5 of 100% =m%. 

(3) $14 = 14 times ^ =8%. 
.-. $14 is 8% of $175. 

PROBLEM 67. 

f is what % of |? 

Solution. 

(1) | = 100%. 

(2) J = i of 100% =-Mp or 33J%. 

(3) | = 4 times 33£% = 133J%. 

(4) f is|ofl33i% = 88*%. 

.-. | is 88| % of f. 

PROBLEM 68. 

30% of the whole of an article is how many % of f of it? 

{R. A., p. 192, prod. 26.) 
Solution. 

(1) 100% = the whole article. 

(2) f of 100% = 66$%, § of the article. 

(3) 66$% = 100%. 

(4) 1% = 100% -i-66|= l.i%. 

(5) 30% = 30. times 1J% = 45%. 

.'. 30% of the whole article is 45% of $ of it. 

PROBLEM 69. 

25% of f of an article is how many % of | of it? 

{R. A., p. 19/, prod. 21.) 
Solution. 



(1) 100% = the whole article. 

(2) -I of 100% = 75%. 

(3) 25% of (| of 100%) = 10%, 

(4) 75% = 100% of itself. 



32 FAIRCHILD'S SOLUTION BOOK. 

(5) 1% =^ of 100% = li%. 

(6) 10% = 10 times 1£% = 13£%. 

.*. 25% of 4 of an article is 13£% of f of it. 

PROBLEM 70. 

If a miller takes 10 quarts of every bushel he grinds for toll, what 
% does he take for toll? {B. N. A., p. 220.) 

Solution. 

(1) 1 bu. = 32 qt. 

(2) 32 qt. = 100%. 

(3) lqt. = J of 100 =31%. 

(4) 10 qt. = 10 times 3£% = 31^%. 

.-. He takes 31£% for toll. 

CASE III. 

PROBLEM 71. 
60 is 20% of what number? 

Solution. 

(1) 100% = number required. 

(2) 20% = 60. 

(3) 1% = 2 V of 60 = 3. 

(4) 100% = 100 times 3 = 300. 

.-. 60 is 20 % of 300. 

PROBLEM 72. 
I is 200% of what number? 

Solution. 

(1) 100% = required number. 

(2) 200% = f. 

(3) 1% = a *T of * = -dhr- 

(4) 100% = 100 times ^ =f|° orf. 

.-. f is 200% of |. 

PROBLEM 73. 

A man owning- 80% of a mill, sold 50% of his share for $8800: at 
this rate, what was the value of the mill? 

Solution. 

(1) 100% = value of mill. 

(2) 80% = his share. 



PER CEN TA GE. 33 

(3) 50% of 80% = 40%, part sold. 

(4) $8800 = value of part sold. 

(5) 40% = $8800. 

(6) 1% = - f V of $8800 = $220. 

(7) 100% = 100 times $220 = $22000. 

.-. $22000 = value of mill. 

PROBLEM 74. 

I pay $13 a month for board, which is 20% of my salary: what is 
my salary? (R. H. A., p. 194, prod. 20.) 

Solution. 

(1) 100% = what I receive per month. 

(2) 20% = $13. 

(3) 1% = Aof$13 = $.65. 

(4) 100% = 100 times $.65 = $65, my monthly salary. 

(5) 12 times $65 = $780, my salary. 

.*. My salary is $780. 

CASE IV. 

PROBLEM 75. 
2576 bu. is 60% less than what? (R. H. A., p. 196, prod. 4.) 

Solution. 

(1) 100% = the number. 

(2) 100% — 60% = 40%, the number decreased by 60%. 

(3) 2567 bu. = the number decreased by 60%. 

(4) .-. 40% = 2576 bu. 

(5) 1% = ^0 of 2576 bu. = 64.4 bu. 

-i%) 100% = 100 times 64.4 bu. = 6440 bu. 

.'. 2576 bu. is 60% less than 6440 bu. 

PROBLEM 76. 

A horse cost $160, which was 20% less than the cost of the carriage: 
what was the cost of the carriag-e? ( Wh. Com., p. 142,) 

Solution. 

(1) 100% = cost of carriage. 

(2) 100% — 20% = 80%, cost of horse. 

(3) $160 = cost of horse. 

(4) 80% = $160. 

(5) 1% = 8 V of $160 = $2. 



34 FAIRCHILD'S SOLUTION BOOK. 

(6) 100% = 100 times $2 = $200. 
.\ The carriage cost $200. 

PROBLEM 77. 

A school enrolls 230 boys, which is 15% more than the number of 
g-irls enrolled: how many pupils in the school? ( Wh. Com., p. 142.) 

Solution. 

(1) 100% = the number of girls. 

(2) 100% + 15% = 115%, the number of boys. 

(3) 230 = number of boys. 

(4) 115% = 230. 

(5) 1% = TJhr of 230 == a 

(6) 100% = 100 times 2 = 200, the number of girls. 

(7) 230 + 200 = 430, the number enrolled. 
.*. There are 430 pupils enrolled. 

PROBLEM 78. 

A coat cost $32; the trimming- cost 70% less, and the making- 50% 
less than the cloth: what did each cost? 

Solution. 

(1) 100% = value of cloth. 

(2) 100% — 70% = 30%, cost of trimming. 

(3) 100% — 50% = 50%, cost of making. 

(4) 100% + 30% + 50% = 180%, cost of coat. 

(5) $32 = cost of coat. 

(6) 180% = $32. 

(7) 1% = T ^of$32 = $.1777*. 

(8) 100% = 100 times $.1777* = $17.77*, cost of cloth. 

(9) 30% = 30 times $.1777* = $5.33J, cost of trimming. 
(10) 50% = 50 times $.1777* =$8.88f, cost of making. 

.-. The cloth cost $17.77*, the trimming $5.33J, the making 

$8 88f . 

PROBLEM 79. 

In a certain company the number of children' was 45% of the num- 
ber of women, the number of women 80% of the number of men, and 
the whole company was 432: how many of each? 

Solution. 

(1) 100% = number of men. 

(2) 80% = number of women. 

(3) 45% of 80 = 36%, number of children. 



PE R CEN PA GE. 35 

(4) 100% + 80% + 36% = 216%, the number in the com- 

pany. 

(5) 432 = number in the company. 

(6) .'. 216 = 432. 

(") 1%=^ of 432= 2. 

(8) 100% = 100 times 2 = 200, the number of men. 

(9) 80% = 80 times 2 = 160, the number of women. 
(10) 36% = 36 times 2 = 72, the number of children. 

.*. There were 200 men, 160 women and 72 children. 

PROBLEM 80. 

Our stock decreased 33J-%, and again 20% ; then it rose 20%, and 
again 33£% ; we have thus lost $66: what was the stock at first? 

Solution. 

(1) 100%? = the original stock. 

(2) 100% — 33J% = 66|%, stock after first decrease. 

(3) 20% of 66J% = 13£%, second decrease. 

(4) 66f%> — 13£% = 53J%, stock after second decrease. 

(5) 20% of 53i % = 10|%, first increase. 

(6) 53J% + 10J% = 64%, stock after first increase. 

(7) 33£% of 64%, = 21J%, second increase. 

(8) 64% + 211 % -= 85J%, stock after second increase. 

(9) 100% — 85J% = 14|%, whole loss. 

(10) $66 = whole loss. 

(11) .-. 14} % = $66. 

(12) 1% = $66 -r 14| = $4 50. 

(13) 100% = 100 times $4.50 = $450, the original stock. 
.'. The original stock was $450. 

PROBLEM 81. 

A brewery is worth 4% less than a tannery, and the tannery 16% 
more than a boat; the owner of the boat has traded it for 75% of the 
brewer}-, losing- thus $103: what is the tanner)- worth? 

Solution. 

(1) 100%, = value of the boat. 

(2) 100% + 16% = 116%, value of the tannery. 

(3) 4% of 116% = 4.64%. 

(4) 116% — 4.64% = 111.36%, value of the brewery. 

(5) 75% of 111.36% = 83.52%, what the owner received for 

the boat. 

(6) 100% —83.52% =16.48%, loss. 



36 FAIRCHILD'S SOLUTION BOOK. 

(7) $103 = loss. 

(8) .'. 16.48% = $103. 

(9) 1% = ^ mTW of $103 = $6.25. 
(10) 116% = 116 times $6.25 = $725. 

.'. The tannery is worth $725. 



CHAPTER V. 

TRADE DISCOUNT. 

30. Discount is the deduction from the list or regular price. 

31. A Net Price is a fixed price from which no discount is 
allowed. 

32. A List Price is an established price, assumed by the sel- 
ler as a basis upon which to calculate discounts. 

33. Trade discounts are often taken off, as "10, 20, 6 and 
5% off," meaning 10% off, 20% off, 6% off and 5% off of the 
remainder. 

PROBLEM 82. * 

Sold 20 dozen feather dusters, giving- the purchaser a discount of 
10, 10 and 10% , his discounts amounting to $325.20: how much was 
my price per dozen? (/?. jd p., p. 209, prod. 5.) 

Solution. 

(1) 100% == my price per dozen. 

(2) 10% of 100% = 10%, 1st discount. 

(3) 100% — 10% = 90%. 

(4) 10% of 90% = 9.%, 2d discount. 

(5) 90% — 9% =81%. 

(6) 10% of 81% = 8.1%, 3d discount. 

(7) 10% + 9% +8.1% = 27.1%, amount of discounts. 

(8) $325.20 = amount of discounts. 

(9) 27.1% = $325.20. 



TRADE DISCOUNT. 37 



(10) 1% = 2I.1 of $325.20 = $12. 

(11) 100% = 100 times $12 = $1200. 

(12) 20 dozen = $1200. 

(13) 1 dozen = ^ of $ 1200 = 
.'. The price per box is 



PROBLEM 83. 
Bought 100 dozen stay bindings at 60 cents per dozen; for 40, 10 and 
!{■%• off: what did I pay for them? 

Solution. 

(1) 60 cents = list price of 1 dozen. 

(2) 100 dozen = 100 times $.60 = $60, list price of 100 dozen. 

(3) 40% of $60 = $24, first discount. 

(4) $60 — $24 = $36, first net proceeds. 

(5) 10% of $36 =$3.60, second discount. 

(6) $36 — $3.60 = $32.40, second net proceeds. 

(7) 1\°/o of $32.40 = $2.43, third discount. 

(8) $32.40 — $2.43 = $29.97, cost. 

.-. I paid $29.97. 

PROBLEM 84. - 

A retail dealer buys a case of slates containing 10 dozen for $50 
list, and gets 50, 10 and 10 off; paying for them in the usual time, he 
gets an additional 2% : what did he pay per dozen for the slates? 

Solution. 

(1) $50 = the list. 

(2) 50% of $50 = $25, first discount. 

(3) $50 — $25 = $25, first net proceeds. 

(4) 10% of $25 = $2.50, second discount. 

(5) $25 — $2.50 = $22.50, second net proceeds. 

(6) 10% of $22.50 = $2.25, third discount. 

(7) $22.50 — $2.25 = $20.25, third net proceeds. 

(8) 2% of $20.25 = $.405, fourth discount. 

(9) $20.25 — $.405 = $19,845, cost of the 10 dozen slates. 
(10) 1 dozen = ^ of $19,845 = $1.9845. 

.'. The slates cost $1.9845 a dozen. 

PROBLEM 85. 

A bookseller purchases books from the publisher at 20% off the list 
price; if he retail them at the list price, what will be his per cent of 
profit? 



38 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) 100% = list price. 

(2) 20% = discount. 

(3) 100% — 20% = 80%, cost. 

(4) 100% = bookseller's selling price. 

(5) 100% — 80% = 20%, gain. 

(6) 80% =20%. 

(7) 1% = io of 20% = i%. 

(8) 100% = 100 times \% = 25%. 

.'. The profit is 25%. 

PROBLEM 86. 

Sold a case of hats containing- 3 dozen, on which I had received a 
discount of 10%, and made a profit of 12£%, or 37^ cents on each hat: 
what was the wholesale merchant's price per case? 

Solution. 

(1) 37-J- cents = profit on one hat. 

(2) $.37| X 36 = $13.50, profit on 3 dozen hats. 

(3) 100% = wholesale merchant's price per case. 

(4) 10% = discount. 

(5) 100% — 10% = 90%, cost. 

(6) 12|% of 90% = lli%, profit. 

(7) 11^% = $13 50. 

(8) 1% = $13 50 -=- 11^ = $1.20. 

(9) 100% = 100 times $1.20= $120, wholesale merchant's 

price per case. 

.*. The price per case was $120. 

PROBLEM 87. 

A dealer in notions buys 60 gross shoestring-s at 70 cents per gross, 
list, 50, 10 and 5% off; if he sells them at 20, 10 and 5% off list, what 
will be his profit? 

Solution. 

(1) 70 cents = list price per gross. 

(2) $.70 X 60 = $42, list price of 60 gross. 

(3) 50% of $42 = $21, 1st discount. 

(4) $42 — $21 = $21, 1st net proceeds. 

(5) 10% of $21 = $2.10, 2d discount. 

(6) $21 — $2.10 = $1890, 2d net proceeds. 

(7) 5% of $18.90 = $.945, 3d discount. 

(8) $18.90 — $.945 = $17,955, cost. 



TRADE DISCOUNT. 39 

Now by 2d condition — 

(9) 20% of $42 = $8.40, 1st dis. by 2d condition. 

(10) $42 — $8.40 = $33.60, 1st net pro. by 2d condition. 

(11) 10% of $33.60 = $3.36, 2d dis. by 2d condition. 

(12) $33.60 — $3.36 = $30.24, 2d net pro. by 2d condition. 

(13) 5% of $30.24 = $1,512, 3d dis. by 2d condition. 

(14) $30.24 — $1,512 = $28,728, selling price. 

(15) $28,728 — $17,955 = $10,773, his profit. 

.-. $10,773 = his profit. 

PROBLEM 88. 

Bought 50 gross of rubber buttons for 25, 10 and 5% off; disposed 
of the lot for $35.91, at a profit of 12% : what was the list price of the 
buttons per gross? 

Solution. 

(1) 100% = list price. 

(2) 25% of 100% = 25%, 1st discount. 

(3) 100% — 25% = 75%, 1st net proceeds. 

(4) 10% of 75% = 7-£%, 2d dis. 

(5) 75% — 7|% = 67|%, 2d net pro. 

(6) 5% of 67|% = 3.375%, 3d dis. 

(7) 67|% — 3.375% = 64.125%, cost. 

(8) 12% of 64.125% = 7.595%, gain. 

(9) 64.125% + 7.695% = 71.82%, selling price. 

(10) $35.91 = selling price. 

(11) 71.82% = $35.91. 

(12) 1% = TT V? of $35.91 = $.50. 

(13) 100% = 100 times $.50 = $50, list price of 50 gross. 

(14) #50 H- 50 = $1, list price of 1 gross. 

.*. $1 = list price per gross. 

PROBLEM 89. 

A merchant imported wine at $2.80 per gal.; 9% was lost by leak- 
age: at what price per gallon must he sell the remainder to gain 30% 
on the cost of all? {Putnam Co.) 

Solution, 

(1) Let 100 = the number of gal. bought. 

(2) 100 X $2.80 = #28, the cost of the wine. 

(3) 30% of $280 = $84, gain. $280 + $84 = $364, selling price. 

(4) 9% of 100 gal. = 9 gal., amount lost by leakage. 

(5) 100 gal. — 9 gal. = 91 gal., remaining. 



40 FAIRCHILD'S SOLUTION BOOK. 

.'. $364 -7- 91 == $4, price at which he sells the remainder 
per gallon in order to gain 30% on the cost of all. 

PROBLEM 90. 

A merchant buys hats from the manufacturers at 25% off list; if he 
retails them at list, what will be his % of profit? 

Solution. 

(1) 100% = list price. 

(2) 25% = discount. 100—25% = 75%, cost. 

(3) 100% = merchant's selling price. 

(4) .'] 100% — 75% = 25%, gain. 

(5) 75% = 100%. 

(6) 1% = T V of 100% = \\% and 25% = 25 times 1£% = 

33i%, gain. 

.*. 33^% is his per cent of profit. 



CHAPTER VI. 



PROFIT AND LOSS. 



34. Profit and Loss are terms which denote the gain or loss 
in business transactions. 

35. Cost is the price paid for goods. 

36. The Selling Price is the price received for goods. 

37. The Profit is what the goods sell for more than they 
cost. 

38. The Loss is what the goods sell for less than they cost. 

PROBLEM 91. 

Sold silk at $1.35 per yard and lost 10% : at what price per yard 
would I have sold it to make a profit of 16|%? 



PROFIT AND LOSS. 41 



Solution. 



I. (1) 100% = cost price; 10% = loss. 

(2) 90% = selling price; $1.35 = selling price. 

(3) 90% = $1.35. 

(4) 1 % = -^ of $1.35 = $ 015. 

(5) 100% = 100 times $.015 = $1.50, cost. 
II. (1) 100% =$150. 

(2) 1% = ^ of $1.50 = $.015. 

(3) 16|% = 16f times $ 015 = $ 25, gain. 

(4) $1.50 + $.25 = $1.75. 

.'. $1.75 = price sold to gain 16J %. 

PROBLEM 92. 

A bookseller sells a grammar for $1.00 which costs 80 cents: what 
is his gain per cent? 

Solution. 

(1) 80 cents = cost price. 

(2) $1.00 = selling price. 

(3) $1.00 — $.80 = $.20, gain. 

(4) 80 cents = 100%. 

(5) 1 cent = -gL f 100% = £%. 

(6) 20 cents = 20 times |% = 25%. 

.'. 25% = his gain. 

PROBLEM 93. 

A merchant sold velvet at a profit of $3 per yard and gained 20% : 
how much did it cost? 

Solution. 

(1) 100% = cost per yard. 

(2) 20% =gain; $3 = gain. 

(3) 20% = $3. 

(4) 1% = ^ of #3 = $.15. 

(5) 100% = 100 times $.15 = $15, price of velvet. 

.'. $15 = cost of velvet. 

PROBLEM 94. 

A merchant sells goods at retail 30% above cost and at wholesale 
12% less than the retail price: what is his gain per cent on goods sold 
at wholesale? ( Wh., p. 148.) 



42 • FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) 100% = cost 

(2) 130% = retail. 

(3) 12% of 130% = 15.6%. 

(4) 130% — 15.6% = 114.4%, wholesale price. 

(5) 114.4% — 100% = 14.4%, gain at wholesale. 

.'. He gained 14.4% at wholesale. 

PROBLEM 95. 

A butcher sold two beeves for $150 each; on the one he gained 25% 
and on the other he lost 25% : did he gain or lose and how much? 

Solution. 

(1) 100% = cost of the first. 

(2) 125% = selling price of first. 

(3) $150 = selling price of first. 

(4) 125% = $150. 

(5) 1% = ^ of $150 = $1.20. 

(6) 100% = 100 times $1.20 = $120, cost of first. 

(7) $150 — $120 = $30, gain on the first. 

(8) 100%, = cost of the second. 

(9) 75% = selling price of the second. 

(10) $150 = selling price of second. 

(11) 75% = $150. 

(12) 1% = T V of $150 = $2. 

(13) 100% = 100 times $2 = $200, cost of the second. 

(14) $200 — $150 = $50, loss. 

(15) $50 — $30 = $20, loss. 

.'. He lost $20 in the transaction. 

PROBLEM 96. 

A merchant reduced the price of cloth 5 cents per yard, and there- 
by reduced his profit on the cloth from 10% to 8% : what was the cost 
of the cloth per yard? 

Solution. 

(1) 100% = cost. 

(2) 10% = profit before deducting. 

(3) 8% = profit after deducting. 

(4) 10% — 8% = 2%, deduction. 

(5) 5 cents = deduction. 

(6) 2% = 5 cents. 



PROFIT AND LOSS. 43 



$2.50. 





(7) 

(8) 


1% = 
100% 


| of 5 cents 
= 100 times 


-f 


: -f cents, 
cents = 


.-. The 


cost price was $2.50. 







PROBLEM 97. 

Sold f of an acre of land for $2 more than § of an acre cost, and 
made 15% on the part sold: find the cost of one acre. 

Solution. 

(1) 100% = cost of one acre. 

(2) f of 100% = 75%, the cost of f of an acre. 

(3) 75% + $2 = the selling price off of an acre. 

(4) (75% + $2) -f- \ = 112|% + $3, selling price of one 

acre. 

(5) Then, by the condition of the problem, 115% = selling 

price of one acre. 

(6) 115% = 112*% + $3. 

(7) 2*% = $3. 

(8) 1% = $3^2^ = 11.20. 

(9) 100% = 100 times $1.20 = $120, cost of one acre. 

.*. $120 = cost of one acre. 

PROBLEM 98. 

A sold two wagons for $150. He gained 5% on the first and 6% on 
the second: how much did each cost if the second cost $10 more than 
the first? 

Solution. 

(1) 100%, = cost of first wagon. 

(2) 5% = gain. 

(3) 100% + 5% = 105%, selling price of first. 

(4) 100% + $10 = cost of 2d wagon. 

(5) 6% of (100% + $10)= 6% + $.60, gain. 

(6) (100% + $10) + (6% + $.60) = 106% + $10.60, selling 

price of 2d. 

(7) 105% + (106% + $10.60) = selling price of both. 

(8) $150 = selling price of both wagons. 

(9) 105% +(106% + $10.60)= $150. 

(10) 211% = $139 40. 

(11) 1% = .,], of $139.40 = $.66^. 

(12) 100% = 100 times $.66^ == $66-J 1 4 i , cost of 1st wagon, 

(13) $66^ + $10 = KTO./.V cost of 2d wagon. 



44 FAIRCHILD'S SOLUTION BOOK. 

.'. The 1st wagon cost $66^, the 2d $76 ir 1 T 4 T . 

Note. — The above is taken from a Wood county examination list, 
and it seemed to bother many old teachers. 

PROBLEM 99. 

A merchant marked a piece of carpet at 25% above cost, and then 
sold it at 20% less than marked price: did he gain or lose, and how 
much? 

Solution. 

(1) 100% = the cost. 

(2) 125% = the marked price. 

(3) 20% of 125% =25%. 

(4) 125% — 25% = 100%, selling price. 

(5) 100% — 100% = 0. 

.'. He neither gained nor lost. 

PROBLEM 100. 

I sold a team of horses for $700; on one I gained 20% , on the other 
I lost 20% ; my total loss was $50: what was the cost of each? 

Solution. 

(1) 100% o = cost of 1st horse, and 20% the gain. 

(2) 120% = selling price of 1st horse. 

(3) $750 — 100% = cost of 2d horse. 

(4) 20% of (1750 — 100%) = 1150 — 20%, loss. 

(5) (1750 — 100%) — ($150 — 20%) = $600 — 80%, selling 

price of 2d horse. 

(6) 120% + ($600 — 80%) = $700. 

(7) 40% = $100. 

(8) 1% = A of $100 = |2.50. 

(9) 100% = 100 times $2.50 = $250, cost of 1st. 
(10) $750 — $250 = $500, cost of 2d. 

.'. The 1st horse cost $220, the 2d $500. 

PROBLEM 101. 

Bought hams at 8 ct. a lb.; the wastage is 10%: how must I sell 
them to gain 30% ? {R. H. A., p. 200, prod. 9.) 

Solution. 

(1) Let 100 = the number of pounds the hams weigh. 

(2) 8 ct. = the cost of 1 pound. 

(3) 100 lb. at 8 ct. = $8, cost. 

(4) 10% of 100 lb. = 10 lb. 



PROFIT AND LOSS. 45 

(5) 100 lb. — 10 lb. = 90 lb., remaining after wastage. 

(6) 30% of |8 = $2.40, gain. 

(7) $8 + 12.40 = $10.40, selling price. 

(8) 90 lb. = $10.40. 

(9) 1 lb. = ^ - of $10.40 = llf cents. 

.\ I sell them at llf cents per pound. 

PROBLEM 102. 

How must cloth costing- $4 a yard be marked that a merchant may 
deduct 15% from the marked price and still make 15% profit? 

Solution. 
I. (1) $4 = cost price. 

(2) 15% = gain. 

(3) 15% of $4 = $.60, gain. 

(4) $4 + $.60 = $4.60, selling price. 
II. (1) 100% = marked price. 

(2) 15% = deduction. 

(3) 100% — 15% = 85%, selling price. 

(4) 85% = $4.60. 

(5) 1% = - 8 V of $4.60 = $.05 T V 

(6) 100% = 100 times $.05-^ = $5y T . 

.*. The marked price is $5 T 7 T . 

PROBLEM 103. 

A sold two horses for $432; on the gray he gained 20% and on the 
bay he lost 20% : did he gain or lose, if % of the cost of the gray equals 
% of the cost of the bay? 

Solution. 

(1) | of the cost of the gray = f of the cost of the bay. 

(2) \ of the cost of the gray = \ of £ = ^ of the cost of the 

bay. 

(3) f the cost of the gray = 3 times ^ of the cost of the 

bay = \l of the cost of the bay (cost of the gray in 
terms of the bay.) 

(4) \l = the cost of the gray. 

(5) -J-ft = tne cost °f tne Da y» 

(6) 20% or i of j * = ft, gain. 

(7) \l + U = ih selling price of the gray. 

(8) 20% or i of « = l, loss. 

(9) 1 o — i = A» selling price of the bay. 

(10) U + A = W, selling price of both. 



46 FAIRCHILD'S SOLUTION BOOK. 

(11) $432 = selling price of both. 

(12) -VV- = $432. 

(13) -fa = T A of 1432 = $3f 

(14) U (H cost of gray) = 60 times $3f = $231f , cost of the 

gray. 

(15) £$ (f§- cost of bay) = 50 times $3f = $192f, cost of the 

bay. 

(16) $231f + $192$ = $424f , amount both cost. 

(17) $432 — $424f = $7f, gain. 

.-. He gained $7f . 

PROBLEM 104. 

Sold two cows for $210. On the first I gained 25% , and on the sec- 
ond I lost 25%: what did I gain if the second cost f as much as the 
first? 

Solution. 

(1) Let j-§ represent the cost of first cow. 

(2) f of -j-f = A» the cost of second cow. 

(3) 25% of if = A, gain. 

(4) -rf + A = tt» selling price of first cow. 

(5) 25% of A = A* loss - 

(6) A — A — A' selling price of second cow. 

(7) tt 4" A = H» selling price of both cows. 

(8) $210 = selling price of both cows. 

(9) fi - $210. 

(10) A = 2V of $210 = $10. 

(11) if = 12 times $10 = $120, cost of first cow. 

(12) A = 8 times $10 = $80, cost of second cow. 

(13) $120 + $80 = $200, cost of both cows. 

(14) $210 — $200 = $10, gain. 

.-. I gained $10. 

PROBLEM 105. 

A stock of g-oods is marked 22 £ % advance of cost, but becoming" 
damag-ed is sold at 20% discount on marked price, whereby a loss of 
$1186.40 is sustained: what was the cost of the g-oods? 

Solution. 

(1) 100% = cost of the goods. 

(2) 122|% = marked price. 

(3) 20% of 122|% = 24^%, discount. 

. (4) 122!% — 24 i% = 98%, selling price. 



PR OFI T A ND L OSS. 47 

(5) 100% — 98% = 2%, loss. 

(6) 81186 40 = loss. 

(7) 2% = 81186.40. 

(8) 1% = J of 11186.40 = 8593.20. 

(9) 100% = 100 times 8593.20 = 859320, cost of the goods. 

.*. 859320 = cost of the goods. 

PROBLEM 106. 

I sold two horses, receiving- 14f % less for the second than for the 
first; on the first I gained 25% and on the second I lost 20% ; my whole 
loss was So: find the cost. 

Solution. 

(1) 100% = selling price of 1st horse. 

(2) 100% = cost of 1st horse. 

(3) 125% = selling price of 1st horse. 

(4) 125% = 100%. 

(5) 1% = ^of 100% =|%. 

(6) 100% = 100 times }% = 80%, cost of 1st horse. 

(7) 100% — 80% = 20%, 'gain. 

(8) 100% = cost of 2d horse. 

(9) 100% — 14} % = 85}%, selling price of 2d. 

(10) 20% of 100% = 20%, loss. 

(11) 100% — 20% = 80%, selling price. 

(12) 80% = 85}%. 

(13) 1% = A of 85}% = 1.06§%. 

(14) 100% = 100 times 1.06}% = 106}%, cost of 2d. 

(15) 106}% — 85}% = 21}%, loss on 2d. 

(16) 21}% — 20% = 1}%, whole loss. 

(17) 85 = whole loss. 

(18) 1}% = 85. 

(19) 1% = 85 + l.\ = 83.75. 

(20) 80% = 80 times 83.75 = 8300, cost of 1st. 

(21) 85}% = 85} times 83.75 = $400, cost of 2d. 

.-. The 1st horse cost 8300, the 2d $400. 

PROBLEM 107. 

How must I mark goods costing $8 per yard that I may deduct 40% 
from the asking price and still make 20%? 

Solution. 
I. (1) 88 = the cost, and 20% the gain. 



48 FAIRCHILD'S SOLUTION BOOK. 

(2) 20% of $8 = $1.60, gain. 

(3) $8 + $1.60 =$9.60, selling price. 
II. (1) 100% = asking price. 

(2) 40% = the reduction. 

(3) 100% —40% = 60%, selling price. 

(4) 60% = $9.60. 

(5) 1% = A of $9.60 = $.16. 

(6) 100% = 100 times $.16 = $16. 

.*. I must mark it $16. 

PROBLEM 108. 

Sold hogs at 10% above cost. Invested $150 more than these pro- 
ceeds in cattle which I sold at 20% loss. If I now have $84 more than 
at first, find value of hogs? 

Solution. 

(1) 100% = cost of the hogs. 

(2) 110% == selling price of the hogs. 

(3) 110% + $150 = cost of the cattle. 

(4) 20% of (110 + $150) = 22% + $30 = loss. 

(5) (110% + $150) — (22% + $30) = 88% + $120. 

(6) 100% + $84 = 88% + $120. 

(7) 12% = $36. 

(8) 1% = T V of $36 = $3. 

(9) 100% - 100 times $3 = $300. 

.-. $300 = value of hogs. 

PROBLEM 109. 
I sold a horse at a gain of 20%, and with this money I bought 
another, which was sold for $166.50, losing 7-| % : find cost of first horse? 

Solution. 

(1) 100% = cost of 1st horse. 

(2) 20% = the gain. 

(3) 120% = selling price of 1st. 

(4) 100% = cost of 2d horse. 

(5) 7|% = the loss. 

(6) 100% — 7-i% = 92-1%, selling price of 2d. 

(7) $166.50 = selling price of 2d. 

(8) 92i% = $166.50. 

"(9) 1% = $166.50 -r- 92i = $1.80. 

(10) 100% = 100 times $1.80 = $180, cost of 2d, also selling 
price of 1st. 



PROFIT AND LOSS. 49 

(11) 120% = $180. 

(12) 1% = T |o of 1180 = 1150. 

(13) 100% = 100 times $1.50 = $150, cost of first horse. 

.'. $150 = cost of first horse. 

PROBLEM 110. 

If a cabinet maker sold a set of furniture for $18.75 more than 
cost, and g-ained 30%, what would have been his rate per cent of gain 
or loss if he had sold the furniture for $87.50? 

Solution. 

(1) 100% = cost of furniture. 

(2) 30% = the advance. 

(3) $18.75 = the advance. 

(4) 30% = $18.75. 

(5) 1% = 3V of $18.75 = $.625. 

(6) 100% = 100 times $.625 = $62.50, cost of furniture. 

(7) $87.50— $62.50 = $25, gain. 

(8) $62 50 = 100%. 

(9) $1 = 100% -=- 62.50 = lf%. 

(10) $25 = 25 times 1|% = 40%, gain. 

.". His rate of gain would have been 40%. 

PROBLEM 111. 

Sold an equal quantity of two kinds of cloth, losing- 3% on the first 
and gaining 5% on the second; the difference in the amount of sales 
was $32.25: find total amount of sales if the first cost \ as much as the 
second. 

Solution. 

(1) 100% = cost of the 2d. 

(2) I of 100% = 75%, cost of the 1st. 

(3) 3% of 75% = 2.25%, loss. 

(4) 75% — 2.25% = 72|%, selling price of 1st. 

(5) 5% of 100% = 5%, gain. 

(6) 100% + 5% = 105%, selling price of 2d. 

(7) 105% — 72|% = 32^%, difference of sales. 

(8) $32.25 = difference of sales. 

(9) 32-i-% = $32.25. 

(10) 1% = $32.25 — 32] = $1. 

(11) 72f % = 72f times $1 = $72f, selling price of 1st. 

(12) 105% = 1<)5 times $1 = $105, selling price of 2d. 



50 FAIR CHILD'S SOLUTION BOOK. 

(13) $72.75 + $105 = $177.75, total amount of sales. 
.'. $177.75 = total amount of sales. 

PROBLEM 112. 

Each of two men, A and B, desired to sell his horse to C. A asked 
a certain price and B 50% more. A then reduced his price 20%, and B 
his price 30% , at which C took both horses, paying- $150: what was 
each man's price? 

Solution. 

(1) 100% = A's asking price. 

(2) 150% = B's asking price. 

(3) 20% of 100% = 20%, reduction. 

(4) 100% — 20% = 80%, A's selling price. 

(5) 30% of 150% = 45%, reduction. 

(6) 150% — 45% = 105%, B's selling price. 

(7) 105% + 80% = 185%, whole selling price. 

(8) $150 = whole selling price. 

(9) 185% = $150. 

(10) 1% = T -b of $150 = $.81 JV. 

(11) 100% = 100 times $.81^ = $81^, A's asking price. 

(12) 150% = 150 times $.81-^ = $121.62^, B's asking price. 

.-. A's asking price is $81^-, and B's is $121.623 6 T . 

PROBLEM 113. 

The sale of a horse was 20% more than that of a cow. My whole 
g-ain was $12. Find cost of each, if I g-ained 25% on the horse and lost 
16f % on the cow. 

Solution. 

(1) 100% = selling price'bf the cow. 

(2) 120% = selling price of the horse. 

(3) 100% = cost of cow. 

(4) 100% — 16f % = 83J%, selling price of cow. 

(5) 83J% = 100%. 

(6) 1% = 100% -^ 83 i = 1.20%. 

(7) 100% = 100 times 1.20% = 120%, cost of the cow. 

(8) 120% — 100% = 20%, loss on the cow. 

(9) 100% = cost of the horse. 

(10) 100% + 25% = 125%, selling price of horse. 

(11) 125% = 120%. 

(12) 1% = T froil20% =U%. 



PROFIT AND LOSS. 51 

(13) 100% = 100 times ff % = 96%, cost of the horse. 

(14) 120% — 96% = 24%, gain on the horse. 

(15) 24% —20% = 4%, gain. 

(16) $12 = gain. 

(17) 4% = $12. 

(18) 1% = \ of |12 = $3. 

(19) 120% = 120 times $3 = $360, cost of cow. 

(20) 96% = 96 times $3 = $288, cost of horse. 
.". The cow cost $360, and the horse $288. 

PROBLEM 114. 

Sold a cow for $69, gaining 15% ; sold another cow for $36, and lost 
the same amount of money as gained upon the first: what was the 
rate of loss on last sale? 

Solution. 

(1) 100% = cost of 1st cow. 

(2) 15% of 100% = 15%, gain. 

(3) 100% + 15% = 115%, selling price of 1st cow. 

(4) $69 = selling price of 1st cow. 

(5) 115% = $69. 

(6) 1% = T H of $69 = $.60. 

(7) 100% = 100 times $.60 = $60, cost of 1st cow. 

(8) $69 — $60 = $9, gain on 1st cow. 

(9) 100% = cost of 2d cow. 

(10) $36 + $9 = $45, cost of 2d cow. 

(11) $45 = 100%. 

(12) $1 = ^ of 100% = 2|%. 

(13) $9 = 9 times 2*% = 20%, rate of loss on last sale. 
.*. 20% was the rate of loss on last sale. 

PROBLEM li:. 

A man bought a horse for a certain price. Now, if he sells him for 
$24, he will lose as much per cent as the horse cost: required the price 
of the horse. 

Solution, 

(1) Let x denote the price. 

(2) Yc\f\ = tne rate °f l° ss - 

(3) xX IM =lS lloss - 
(4)x-jg=24. 



52 FAIRCHILD'S SOLUTION BOOK. 

(5) x2 — lOOx = —2400. 

(6) x 2 — lOOx + 2500 = 100. 

(7) x — 50 = ±10. 

(8) x = 60 or 40. 

.'. The price of the horse was either $60 or $40. 

PROBLEM 116. 

Bought 400 lb. of tea, and 1600 lb. of sugar. A pound of sugar cost 
me \ as much as a pound of tea. Sold the tea at a gain of 33^% , and 
the sugar at a loss of 20%. Find the investment, if my entire gain is 
$60. 

Solution. 

(1) 100% = cost of tea. 

(2) } of 100% = 16} %, cost of same no. lb. sugar as tea. 

(3) 4 times 16}% = 66}%, entire cost of sugar. 

(4) 33-1;% of 100% = 33^%, g a i n on tea. 

(5) 20% of 66}% = 13J%, loss on sugar. 

(6) 33}% — 13J% = 20%, entire gain. 

(7) $60 = entire gain. 

(8) 20% = $60. 

(9) 1% =Jf of $60 = $3. 

(10) 100% = 100 times $3 = $300, cost of tea. 

(11) 66}% = 66} times $3 = $200, cost of sugar. 

(12) $300 + $200 = $500, the entire cost. 

.'. $500 is the investment. 

PROBLEM 117. 

A drover sold a cow for $25, losing 16f % ; bought another and sold 
her at a gain af 16%. He neither gained nor lost on the two: find 
their cost. 

Solution. 

(1) 100% = cost of 1st cow. 

(2) 100% — 16} % = 83}%, selling price of 1st cow. 

(3) $25 = selling price of 1st cow. 

(4) 83}% = $25. 

(5) 1% = $25 -=- 83} = $.30. 

(6) 100% = 100 times $.30 = $30, cost of 1st cow. 

(7) $30 — $25 = $5, loss on 1st cow. 

(8) 100% = cost of second cow. 

(9) Now, as he neither gained nor lost on the two, he must 

buy the second so as to gain $5. 



PROFIT AND LOSS. 53 

(10) Then, we can readily see that 16% = $5. 

(11) 1% = T ig- of $5 = $.3125. 

(12) 100% = 100 times 1.3125 = $31.25. 

.-. The 1st cow cost $30, and the 2d $31.25. 

PROBLEM 118. 
The cost of transporting- goods is 9% of their cost; if a merchant 
wishes to make a profit of 25%, how must he mark his goods? What 
amount of goods must he bti}' to make a profit of $3625? 

Solution. 

(1) 100% = cost of goods. 

(2) 9% of 100% = 9%, cost of transportation. 

(3) 100% + 9% = 109%, cost of goods after transportation. 

(4) 25% of 109% = 27i%, gain. 

(5) /. He must mark his goods 27£% + 9% = 36J% above the 

1st cost. 

(6) 36|% — $3625. 

(7) 1% = $3625 ~ 36J% = $100. 

(8) 100% = 100 times $100 = $10000, cost of the goods. 

.'. The goods cost $10000 and must be sold 36j-% above 
cost to gain $3625. 

PROBLEM 119. 

Bought 3 cows for $120, and sold them at equal prices. On the 
first I gained 60'/ , on the second 20% , and on the third I lost \ c /c : what 
did each cost? 

Solution. 
100% = selling price of each cow . . . (A). 
I. (1) 100% = cost of 1st cow. 

(2) 60% = gain on 1st cow. 

(3) 100% + 60% = 160%, selling price of first cow. 

(4) 160% = 100% ... (A). 

(5) 1% = ^ of 100% = £%. 

(6) 100% = 100 times |% = 62|%, cost of 1st cow. 
II. (1) 100% = cost of 2d cow. 

(2) 20% = gain on 2d cow. 

(3) 100% + 20% = 1207c, selling price of 2d cow. 

(4) 120% = 100% . . . (A). 

(5) 1% = ^ of 100% =|-%. 

(6) 100% = 100 times •>% = 83J%, cost of 2d cow. 
III. (1) 100% = cost of 3d cow. 



54 FAIRCHILD'S SOLUTION BOOK. 

(2) 4% = loss on 3d cow. 

(3) 100% — 4% = 96%, selling price of 3d cow. 

(4) 96% = 100% ... (A). 

(5) 1% = A- of 100% =|f%. 

(6) 100% = 100 times |f% = 104^%, cost of 3d cow. 
IV. (1) 62i% _|_ 83^% + 1041% = 250%, cost of the three 

cows. 

(2) $120 = cost of the three cows. 

(3) 250% = $120. 

(4) 1% =Tirr of $120 = $.48. 

(5) 62i% - 62i times $.48 = $30, cost of 1st cow. 

(6) 83i% = 83^ times $.48 = $40, cost of 2d cow. 

(7) 104i% = 1041 times $.48 = $50, cost of 3d cow. 

.-. The 1st cow cost $30, the 2d $40, and the 3d 



PROBLEM 120. 

A merchant sold a hat for $2; his cost mark was "nek," and his 
key "now be quick": find his gain per cent. 

Solution. 

(1) By the key n = 1, e = 5 and k = 0. 

(2) 100% = the cost. 

(3) $2 — $150 = $.50, gain. 

(4) $1.50 = 100%. 

(5) $1 = T ,b of 100% = 66|%. 

(6) $.50 = .50 times 66|% = 33^%, rate of gain. 

.-. The rate of gain is 33^%. 
Note. — You can readily see that "nek" == 150. 

PROBLEM 121. 

A merchant sells goods at 50% profit and takes eggs at market 
price in payment: if one egg in each dozen is bad, find his rate of net 
gain. 

Solution. 

(1) 100% = cost of goods. 

(2) 50% == gain. 

(3) 150% = selling price of the goods. 

(4) By the last condition he loses ^ of his selling price. 

(5) Then, ft of 150% = 12i%, loss. 

(6) 50% — 12i% = 374%, his net S ain - 
.*. His net gain is 37^%. 



PROFIT AND LOSS. 55 

PROBLEM 122. 

An implement dealer asked for a reaper 30% more than it cost him; 
if he finally took 20% less than his asking- price, and gained $5 on the 
machine, find the cost. 

Solution. 

(1) 100% = the cost of the implement. 

(2) 100% + 30% = 130%, asking price. 

(3) 20% of 130% =26%. 

(4) 130% — 26% = 104%, selling price. 

(5) 104% — 100% = 4%, gain. 

(6) $5 = gain. 

(7) 4% = $5. 

(8) 1% = J of $5 = $1.25. 

(9) 100% = 100 times $1.25 = $125, cost. 

.'. The implement cost $125. 

PROBLEM 123. 

A cow and horse cost $132: required the cost of each, if the cow- 
cost f as much as the horse, minus $8. 

First Solution. 

(1) 100% = cost of horse. 

(2) Cost of cow = | of 100% = 40%, cost of horse — $8. 

(3) 100% + 40% — $8 = cost of both. 

(4) $132 = cost of both. 

(5) 100% + 40% — $8 = $132. 

(6) 140% = $140. 

(<) 1% = dhr°f$140 = $1. 

(8) 100% = 100 times $1 = $100, cost of horse. 

(9) 40% = 40 times $1 = $40. 
(10) $40 — $8 = $32, cost of cow. 

Second Solution. 

(1) The cost of the horse = f cost of the horse. 

(2) The cost of the cow = f cost of the horse — $8. 

(3) The cost of both = \ cost of the horse — $8. 

(4) .-. I cost of horse — $8 = $132. 

(5) i cost of horse = $140. 

(6) Then ± of cost of horse = i of $140 = $20. 

(7) | the cost of horse = 5 times $20 = $100. 

(8) | cost of horse — $8 (or cost of cow) = $32. 

.-. The horse cost $100, and the cow $32. 



56 FAIRCHIL&S SOLUTION BOOK. 

PROBLEM 124. 

I sold an article for \ more than it cost me to A, who sold it for $6, 
which was f less than it cost him: what did it cost me? 

Solution. 

(1) 100% = cost. 

(2) \ or 25% = my gain. 

(3) 125% = my selling price, also A's cost. 

(4) I- of 125% = 50%. 

(5) 125% — 50% = 75%, A's selling price. 

(6) $6 = A's selling price. 

(7) 75% = $6. 

(8) 1% =^o£W = $.08. 

(9) 100% = 100 times $.08 = $8, what it cost me. 

.*. The article cost me $8. 

PROBLEM 125. 

A bought two horses for $300, and sold them at $205 each, gaining 
20% more on the one than on the other: find the cost of each. 

Solution. 

(1) $300 = cost of both. 

(2) $205 + $205 = $410, selling price of both. 

(3) $410 — $300 = $110, gain on both. 

(4) 100% = gain on 1st horse. 

(5) 120% = gain on 2d horse. 

(6) 100% + 120% = 220%, gain on both. 

(7) |110 = gain on both. 

(8) 220% = $110. 

(9) 1% = ^ of $110 = $.50. 

(10) 100% = 100 times $.50 = $50, gain on 1st horse. 

(11) 120% = 120 times $.50 = $60, gain on 2d horse. 

(12) $205 — $50 = $155, cost of 1st horse. 

(13) $205 — $60 = $145, cost of 2d horse. 

.-. The 1st horse cost $155, and the 2d $145. 

PROBLEM 126. 

I bought an article and sold it so as to gain 10% ; if it had cost 20% 
less, and I had sold it for one dollar less, I would have gained 25% : find 
cost of the article. 



PR OFI T A ND L OSS. 57 

Solution. 

(1) 100% = actual cost. 

(2) 110% = actual selling price. 

(3) 80% = supposed cost. 

(4) 25% of 80% = 20%, gain. 

(5) 80% + 20% = 100%, supposed selling price. 

(6) 110%, actual selling price — 100%, supposed selling price 

= 10%. 

(7) 10% = $1. 

(8) 1% = T V of |1 = $.10. 

(9) 100% = 100 times $.10 = $10, the cost of the article. 

.-. The article cost $10. 

PROBLEM 127. 

A merchant marked goods to gain 60%, but on account of using- an 
incorrect yard-stick, he gained only 30% : find the length of the 
measure. 

Solution. 

(1) 100% = cost. 

(2) 160% = marked price. 

(3) 130% = selling price. 

(4) Hence ||§ = 1^ yards, length of yard-stick. 

.'. The length is 1 T 3 3 yards. 

PROBLEM 128. 

I bought 75 barrels of flour at $9 a barrel, and sold J of it at a cer- 
tain gain per cent, \ at twice that gain, and the remainder at a net 
profit of $25. If I had sold the last lot for $8.75 more, I would have 
gained 10% on the whole: what was my gain per cent on the first lot, 
and my total gain? 

Solution. 

(1) $9 X 75 = $675, the cost of the flour; he sold the last £ 

for $225 + $25 = $250. 

(2) 10% = gain on the whole. 

(3) 10% of $675 = $67.50, gain on the whole. 

(4) $675 + $67.50 = $742.50, what he would have received 

on the whole. 

(5) $742.50 — ($250 + $8.75) = $483.75, the amount he re- 

ceived on the first J. 

(6) $483.75 — $450 = $33.75, gain. 

(7) Let J- be the gain on the 1st; then § is the gain on the 2d. 

(8) i of $33.75 = $11.25, gain on the 1st J. 



58 FAIR CHILD'S SOLUTION BOOK. 

(9) 100% = cost of the 1st J, or $225. 

(10) $225 = 100%. 

(11) $1 = -^ of 100% - i%. 

(12) $11.25 = 11.25 times f % = 5%, rate of gain. 

(13) ($483 75 + $250) — $675 = $58 75, total net gain. 

.*. The required rate is 5%, and the gain $58.75. 

PROBLEM 129. 

A man wishing to sell a horse and a buggy, asked three times as 
much for the horse as for the buggy, but finding no purchaser, he re- 
duced the price of the horse 20% and the price of the buggy 10%, and 
sold them both for $165: what did he get for each? 

Solution. 

(1) 100% = asking price of buggy. 

(2) 300% = asking price of horse. 

(3) 10% of 100% = 10%, deducted from price of buggy. 

(4) 100% — 10% = 90%, selling price of buggy. 

(5) 30% of 300% = 60%, deducted from price of horse. 

(6) 300% — 60% = 240%, selling price of horse. 

(7) 240% + 90% = 330%, selling price of both. 

(8) $165 = selling price of both. 

(9) 330% = $165. 

(10) 1% = 3^ of $165 = $.50. 

(11) 90% = 90 times $.50 = $45, selling price of buggy. 

(12) 240% = 240 times $.50 = $120, selling price of horse. 

.'. He received $45 for the buggy, and $120 for the horse. 

PROBLEM 130. 

A merchant bought sugar at 20% less than its market value, and 
received a discount of 1% for cash; he sold it at an advance of 15% 
above market value: what was his gain per cent? 

Solution. 
I. (1) 100% = market value. 

(2) 80% = price to merchant. 

(3) 4% of 80% = 3.20%, discount. 

(4) 80%? — 3.20% = 76.80%, actual cost to merchant. 
II. (1) 115% = selling price. 

(2) 115% — 76.80% = 38.20%, gain. 

(3) 76.80% = 100% of itself. 

(4) 1% = t ,l ¥ of 100% = -,\%%%. 



PROFIT AND LOSS. 59 

(5) 38 20% = 38.20 times ^\%%= 49f£%, gain per cent. 
.*. His rate of gain was 49-JJ%. 

PROBLEM 131. 

Sold an article at 20% gain; had it cost $300 more, I would have 
lost 20% : find the cost. (R. H.,p. 409.) 

Solution. 

(1) 100% = actual cost. 

(2) 100% + 20% = 120%, actual selling price. 

(3) 100% + $300 = supposed cost. 

(4) 20% = loss on supposed cost. 

(5) 20% of (100% + $300) = 20% + $60, loss. 

(6) (100% + $300) — (20% + $60) = 80% + $240, sell, price. 

(7) 120% = 80% + $240. 

(8) 40% = $240. 

(9) 1% = - 4 V of $240 = $6. 

(10) 100% = 100 times $6 = $600. 

.'. The article cost $600. 

PROBLEM 132. 

I sell goods and gain 20% ; if they had cost me $60 less I would have 
gained 25% : find the cost. 

Solution. 

(1) 100% = actual cost. 

(2) 100% + 20% = actual selling price. 

(3) 100% — $60 = supposed cost. 

(4) 25% == gain on supposed cost. 

(5) 25% of (100% — $60) = 25% — $15, gain. 

(6) (100% — $60) + (25% — $15) = 125% — $75, sellg pr. 

(7) 120% = 125% —$75. 

(8) 5% = $75. 

(9) 1% = l of $75 = $15. 

(10) 100% = 100 times $15 = $1500. 

.*. The goods cost $1500. 

PROBLEM 133. 

If an article had cost me 10% less, the amount of per cent gain 
would have been 15% more: what was the gain? 

First Solution. 

(1) 100%=cost. Then, if the cost were 90% I would gain 15% 
of 90%, or 13V% more. Now, the additional gain of 



60 FAIR CHILD'S SOL V TION BOOK. 

13%% includes the gain of 10% in the reduction of cost, 
and 10% of the cost gains only 13£% — 10% = 3|%. 
(2) If the gain on 10% of the cost is 3|%, 100%, or the cost, 
it is 10 X 3| = 35%, gain. 

Second Solution. 

(1) The selling price divided by the cost gives the amount 

received for $1 of cost. Since the selling price is the 
same in each case, the difference between these 
amounts is the difference in gain per cent; hence — 

( 2 ) 9V — ih> = goo- difference in the selling price. 

(3) rfo = -15%. 

(4) fgg- = 900 times .15% = 135%, selling price. 

(5) 135% — 100%, cost = 35%, gain. 
.*. The gain was 35%. 

PROBLEM 134. 
A man sells two farms for the same price; on the first he gains 
20%, and on the second, he loses 20%, in the aggregate losing $20: 
what did each farm cost? 

Solution. 

100% = selling price of each farm . . . (A). 
I. (1) 100% = cost of 1st farm. 

(2) 20% = gain on the 1st farm. 

(3) 100% + 20% = 120%, selling price of 1st farm. 

(4) 120% = 100% ... (A). 

(5) 1% = T io of 100% = |%. 

(6) 100% = 100 times f% = 83J%, cost of 1st farm. 

(7) 100% — 83^% = 16f %, gain on 1st farm. 
II. (1) 100% = cost of 2d farm. 

(2) 20% = loss on 2d farm. 

(3) 100% — 20% = 80%, selling price of 2d farm. 

(4) 80% = 100% . . . (A). 

(5) 1% = 3V °f 100% = 1^%. 

(6) 100% = 100 times 1^% = 125%, cost of 2d farm. 
III. (1) 125% —100% = 25%, loss on 2d farm. 

(2) 25% — 16f % = 8i%, the whole loss. 

(3) $20 = whole loss. 

(4) 8i% = $20. 

(5) 1% = $20 -f- 8£ = $2.40. 

(6) 83J% = 83i times $2 40 = $200, cost of 1st farm. 

(7) 125% = 125 times $2.40 = $300, cost of 2d farm. 
.-. The 1st farm cost $200 and the 2d $300. 



PROFIT AND LOSS. 61 



PROBLEM 135. 



A merchant sold part of his goods at 25% profit, and the remainder 
at a loss of 15%. His goods cost him $1000, and his gain was $130: 
how much was sold at a profit? 

Solution. 

(1) 100% = cost of goods sold at a profit. 

(2) $1000 — 100% - cost of those sold at a loss. 

(3) 25% = profit on the part sold at a profit. 

(4) 125% = selling price of the part sold at a profit. 

(5) 15% of ($1000 — 100%) = $150 —15%, loss. 

(6) ($1000 — 100%) — ($150 — 15%) = $850 — 85%, sell- 

ing price of part sold at a loss. 

(7) 125% + $850 — 85% = whole selling price. 

(8) $1130 = whole selling price. 

(9) 125% + $850 — 85% = $1130. 

(10) 40% = $280. . 

(11) 1% = ^of$280 = $7. 

(12) 100% = 100 times $7 = $700, cost of goods sold at a gain. 

.'. The goods sold at a gain cost $700. 

PROBLEM 136. 

Bought two cows, paying $300 for one and $450 for the other; sold 
both for the same sum, gaining as many per cent on the first as I lost 
on the second: find the rate of gain or loss. 

Solution. 

(1) 100% = cost of 1st COW. 

(2) r = rate of gain. 

(3) 100% + r = selling price. 

(4) (100% + r) = (100% + r) times $300 = $300 + 300r, 

selling price of 1st cow. 

(5) 100% = cost of 2d cow. 

(6) 100% — r = selling price of 2d cow. 

(7) (100% — r) = (100%— r) times $450 = $450 — 450r, 

selling price of 2d cow. 

(8) Since he sold both for the same sum, $300 -f- 300r = 

$450 — 450r. 

(9) 750r = $150. 

(10) r = r io of $150 = 20 cents, or 20%. 

.*. 20% is the rate of gain or loss. 



62 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 137. 

A man bought two farms for $1000; he sold them for $800 apiece, 
the gain on the one being- 10% more than on the other: what was the 
gain on each? 

Solution. 

(1) Let x = cost of 1st farm. 

(2) $1000 — x = cost of 2d farm. 

(3) $800 — x = gain on 1st farm. 

(4) $800 — ($1000 — x) = gain on 2d farm. 



(5) . ^ v "" = the rate of gain on 1st. 



^ $1000 — x = the mte ° f gain ° n 2d ' 



._ -x x — $200 _ ^ 

{ ° x $1000— x ~~ 10 ' 

(8) (8000000— 18000x+10x 2 ) — (10x 2 — 2000x) = lOOOx— x 2 . 

(9) 8000000 — 16000x = lOOOx — x 2 . 

(10) Transposing, x 2 — 17000x = —8000000, 

(11) Completing square, x 2 — 17000x + 72250000=61250000. 

(12) x — 8500 = ±8015.60+, whence x = $484.40. 

(13) 800 — x = $315.60, gain on 1st. 

(14) x — $200 = $284.40, gain on 2d. 

Note. — This problem was solved, for the Teachers' 1 Review, by A.O. 
Spear. J. W. Watson gives $285f- and $314f for the first and second. 



CHAPTER YII. 



COMMISSION. 



39. Commission is a percentage paid to an agent for the 
transaction of business. 

40. An Agent is a person who transacts business for another; 
he is often called a Commission Merchant, a Factor, etc. 

41. The Principal is the person for whom the commission 
merchant transacts the business. 

42. A Consignment is a quantity of merchandise sent to a 
commission merchant to be sold. 

43. The person sending the merchandise is the Consignor 
or Shipper, and the commission merchant is the Consignee. 

44.- The Net Proceeds is the sum left after the commission 
and charges have been deducted from the amount of a sale or 
collection. 

45. The Entire Cost is the sum obtained by adding the com- 
mission and charges to the amount of a purchase. 

46. A Broker is a person who deals in money, bills of credit, 
stocks, or real estate, etc. 

47. The commission paid to a broker is called Brokerage. 

NoTK. — In this, aud also the next chapter, we are forced to reject 
the "100% method" and use $1 as the basis of computation. Mathema- 
ticians are fast coming- over to the "dollar method." Why not lay the 
sickle on the shelf of some good museum, and let the young minds of 
this great nation reap the golden harvest with the self-binder? 

PROBLEM 138. 

An agent bought a house for $6500, his commission being %\%\ 
what was his commission? 

(63) 



64 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) Out of every $1 invested the agent receives 3-J- ct; and 
for the investment of 16500 he would receive 6500 X 
3| ct. i= $227.50. 

.•. The agent receives $227.50 commission. 

PROBLEM 139. 

My agent bought 40 horses for $150 each, and paid $25 for their 
keeping- and $80 for transportation; his commission was 3£% : what 
did the horses cost me? {Brooks' 1 N. N. W. A., prod, j, p. 22 7.) 

Solution. 

(1) $150 times 40 = $6000, amount paid for the horses., 

(2) Now, out of everv $1 that the agent paid for the horses 

he received 3|'ct.; then, he received $6000 ~ .03-J- = 
$210, commission. 

(3) $210 + $25 + $80 = $315, cost of keeping, transporta- 

tion and commission. 

(4) $6000 + $315 = $6315, whole cost. 

.*. The horses cost me $6315. 

PROBLEM 140. 

A factor sold some land, and paid over $7742.10, retaining- $117.90 as 
commission: required the rate. (Brooks' JV. JV. W. A., prod. 2, p. 229.) 

Solution. 

(1) $7742.10 + $117.90 = $6890, what the land sold for. 

(2) $117.90 = the factor's commission. 

(3) If the factor had received 1 ct. on every $1 worth of land 

sold, he would have received $78.60; then, his rate of 
commission must be as many per cent as $78.60 is con- 
tained times in $117.90, or 1|%. 

.'. His rate of commission is 1\%. 

PROBLEM 141. 

I sold some goods on commission at 5% through an agent, who 
charged me 3% ; my commission, after paying my agent, was $383: re- 
quired the agent's commission, my commission, and the money paid to 
my employers. {Brook's JV. N. W. A., prod, g, p. 226.) 

Solution. 

(1) $338 = money left after paying my agent. 

(2) 5ct. = my commission on each $1 sold; 3ct. = the com- 

mission I pay my agent for selling the same goods. 

(3) 5 ct. — 3 ct. = 2 ct., my money left after paying my 

agent. 



COMMISSION. 65 

(4) .*. There must have been as many dollars worth of goods 

as 2 ct. is contained times in $388, or $19400. 

(5) $19400 X 5 ct. • = $970, my commission. 

(6) $19400 X 3 ct. = $582, my agent's commission. 

(7) $19400 — $970 = $18430, amount remitted to my employ- 

ers. 

.*. $970 is my commission, $580 my agent's commission, 
and $18430 the amount remitted to my employers. 

PROBLEM 142. 

At a commission of 2%, a commission merchant received $8 for 
selling- 24 barrels of cider: for how much per barrel did he sell the 
cider? 

Solution. 

(1) Out of every $1 of receipts for cider, the commission 

agent receives 2 ct. commission. 

(2) $8 is his total commission; therefore, there must have 

been as many dollars in the receipts for cider as 2 ct. 
is contained times in $8, or $400. 

(3) Then -^ of $400, or $16§, is the selling price per bbl. 

PROBLEM 143. 

An attorney collected money and retained $50 commission at 2% : 
required amount and net proceeds. [Schuyler's H. A., prod. 4, p. 316.) 

Solution. 

(1) For every $1 collected, the attorney receives 2 ct. com- 

mission; .'. there must have been as many dollars col- 
lected as 2 ct. is contained times in $50, or $2500. 

(2) Then, $2500 — $50 = $2450, net proceeds. 

.-. The amount is $2500, and the net proceeds $2450. 

PROBLEM 144. 

An ag-ent bought 40 horses for the government at $150 apiece; the 
freight was $160, the agent's commission was such that the horses cost 
the g-overnment $6460: what per cent was the commission? 

Solution. 

(1) $150 X 40 = $6000, cost of the horses; this cost plus the 

commission is $6460 — $160 = $6300. 

(2) Hence, the commission is $300, or $300 ~ $6000 = 5%. 

.*. The commission was 5%. 

Note. — The investment always consists of 100 parts; and the com- 
mission of as many parts as stated in the problem. For example: an 



66 FAIRCHIL&S SOLUTION BOOK. 

ag-ent receives $105 with which to buy hats; after deducting- his com- 
mission of b%, what must be expended? $105 = investment -\- com- 
mission; 100 parts = investment and 5 parts = commission. There- 
fore, if r = his commission, and \%% = his investment; y-fj-g- of $105, or 
$5 = ag-ent's commission, and T #f of $105 = $100, amount expended. 
The student should study these points until they become familiar. If 
you do this you will have but little trouble with the problems on pag-e 
219, Ray's Higher Arithmetic. 

PROBLEM 145. 

Sent my ag-ent $1248 with instructions to buy flour, rate of commis- 
sion 4%: what was his commission, and how many barrels did he buy 
at $6 per barrel? 

Solution. 

(1) Out of every $1 invested in flour the agent receives 4 ct. 

commission. Now, the investment is -J^J, and y^ = 
the amount of commission. 

(2) yfo of $1248 = $48, agent's commission. 

(3) m of $1248 = $1200, amount invested in flour. 

(4) $1200 ~ 6 = 200, no. of bbl. 

.'. The agent's com. is $48, and 200 bbl. were bought. 

PROBLEM 146. 

An ag-ent sold land at 5% commission, and invested the net pro- 
ceeds in wheat at 2% commission; his whole commission was $630: for 
what did he sell the land, and what did he pay for the wheat? 

(Schuyler 's H. A., p. j/6, prob. g.) 

Solution. 

(1) Out of every $1 received for land, the agent received 1st 

5 ct., and had left 95 ct. proceeds to invest in wheat and 
to pay his commission of 2%. 

(2) Now, this 95 ct. contains two parts; 1st, 100 parts invest- 

ment and 2 parts commission; then, T %^ of 95 ct., or 
i|-| ct. = agent's 2d commission. 

(3) .*. Out of every $1 received for land the agent receives 1st 

5 ct., and 2d ||f ct., or ££J ct. = %^. 

(4) But $630 = agent's commission on both transactions; 

hence, there must have been as many dollars received 
for land as $y^ T is contained times in $630, or $9180. 

(5) $9180 — $630 = $8550. 

.-. He received $9180 for the land and invested $8550 in 
wheat. 



COMMISSION. 67 

PROBLEM 147. 

An ag-ent sold cotton at 2|% commission, and invested § of the 
amount of sale in sugar at 2$-% commission; the balance, $1915. 22|-, was 
remitted: what was the value of the cotton, the sug-ar and the whole 
commission? {Schuyler's H. A., p. 317, prod. 10.) 

Solution. 

(1) Out of every $1 received for cotton, the agent keeps 2-|- 

ct. and invests 75 ct. in sugar, keeping also 2-J-% of 75- 
ct., or If ct. 

(2) 2| ct. + 75 ct. + If ct. = 79^ ct., amount kept out of 

every $1 received for cotton. 

(3) $1 — .79J = $.2075, balance out of each $1 remitted to 

principal. 

(4) $1915.221 = amount remitted to principal. .'. There must 

have been as many dollars received for cotton as $.2075 
is contained times in $1915.225, or $9230. 

(5) f of $9230 = $6922.50, amount paid for sugar. 

(6) $9230 X 2\ ct. = $230.75, com. on sugar. 

(7) $6922.50 X 2J ct. = $161,525, com. on sugar. 

.-. $9230 is the value of the cotton; $6922.50, the value of 
the sugar; and $392,275, the whole com. 

PROBLEM 148. 

A man sold a horse, losing 25%; keeping - $60 of the proceeds, he 
gave the remainder to an agent to buy hogs, commission 16| % ; he lost 
in all $62? : what was the value of the horse? 

First Solution. 

(1) For each $1 in the worth of the horse, he received in the 

sale 75 ct. Now, this 75 ct. is made up of 100 parts in- 
vestment and 16$ parts commission; hence, the invest- 
ment was TT^i °f ?5 ct » or 64^ ct. 

(2) Had no money been retained, the loss would have been 

354 ct. on $1 of the original value, i. e. % he would have 
lost {\ of the whole had he left the $60 in the proceeds. 
But the loss on the $60 is simply the commission, 

161 

—— of $60, or $84. Had he kept no money, the whole 
llbj 

loss would have been $621 + $8i = $711; and this 

would have been ^ of the value of the horse. 

(3) Then, & = #71-2. 

(4) ^ = -l of $71 § = $14f 

(5) V| = 14 times $14^ = $200, the value of the horse. 



68 FAIRCHILD'S SOLUTION BOOK. 

Second Solution. 

(1) 100% = the value of the horse. 

(2) 25% = loss. 

(3) 100% — 25% = 75%, selling price of horse. 

16^ 

(4) ^ of (75% — $60) = (10$ % — $8f ) commission. 

(5) 25% + (10$% — $8$) = whole loss. 

(6) $62$ = whole loss. 

(7) 25% + (10$ % — $8$) = $62$, 

(8) 35$ % = $71$. 

(9) 1% = $71$ 4- 35$ = $2. 

(10) 100% = 100 times $2 == $200, value of the horse. 

.-. $200 = the value of the horse. 

PROBLEM 149. 

An agent made $400 by selling- wheat at 4% commission and buying 
cattle with the proceeds, after retaining his commission of 20%: find 
sale of wheat. 

Solution. 

(1) Out of every $1 received for wheat, the agent received 4 

ct. and had left 96 ct. Now, this 96 ct. contains 100 
parts investment and 20 parts commission. 

(2) T \°o of 96 ct. = 16 ct., agent's 2d com. 

(3) Then, out of every $1 received for wheat, the agent re- 

ceives 1st 4 ct., and 2d 16 ct., or 20 ct. 

(4) $400 = agent's com. for both transactions; then, there 

must have been as many dollars received from the sale 
of wheat as $.20 is contained times in $400, or $2000. 

== the sale of wheat. 

PROBLEM 150. 

Sold a consig-nment of pork, and invested the proceeds in brandy, 
after deducting my commissions, 4% for selling, and \\% for buying; 
the brandy cost $2304: what did the pork sell for, and what were my 
commissions? {R. H. A., p. 2ig, prob. 8.) 

Solution. 

(1) Out of every $1 received from the sale of pork, the agent 

receives 4 ct. and has 96 ct. proceeds. Now, this 96 ct. 
is made up of 100 parts investment and 1^ parts com- 
mission. 

(2) Hence, -^— of 96 ct. = $T7yn> amount invested in brandy. 






COMMISSION. 69 

Therefore, there must have been as many dollars re- 

96 
ceived from the sale of pork as ^— is contained times 

in $2304, or $2430. 

(3) $2430 X .04 = $97.20, 1st commission. 

(4) $2304 X .01± = $28.80, 2d commission. 

/. The pork sold for $2430; 1st com., $97.20; 2d com., 
$28.80. 

PROBLEM 151. 

My agent sells pork at 4% commission; after increasing the pro- 
ceeds by $2.60, I order him to purchase wheat at 4% commission; wheat 
now declined 2^% , and my total loss amounts to $7.31^: what did the 
pork bring - ? 

Sohition. 

(1) Out of every $1 received for pork, the agent receives 

4 ct, leaving 96 ct. proceeds. Now, this 96 ct. is made 
up of 100 parts investment and 4 parts commission. 

(2) Hence, T J T of 96 ct. = f-JJ ct., agent's 2d commission. 

(3) i^oV of tU of 96 ct. = f oi ct., loss b y decline. 

(4) In all my loss on one dollar of the receipts for pork, 4 ct. 

+ m ct. + m ct. = $rWo. 

(5) Now, my loss on the $2.60 is first -^ of $2.60 = $Vo¥- 

and next T %% of H$ of $2 60 = $Jfc. 

(6) $W\° + IsW = l 3 lff. total loss on $2.60. 

(7) Now, $7.31J — $ 3 |^° = ^Hi^-y loss on the value of pork. 

(8) If the loss on $1 of the receipts for pork is $ T V^o, then 

there must have been as many dollars in the receipts 
for pork as tVA ls contained times in $ 1 -- 4 |^ 2 - Q -, or $71.50. 

.'. The pork sold for $71.50. 

PROBLEM 152. 

My agent sold cattle at 10% commission, and after I increased the 
proceeds by $18 I ordered him to buy hogs at 20% commission; the hogs 
had declined 6|%, when he sold them at 14f % commission; I lost $86 in 
all: for what did the cattle sell? 

Solution. 

(1) Out of every $1 received for cattle the agent receives 10 

ct., leaving 90 ct. proceeds. Now, this 90 ct. is made 
up of 100 parts investment and 20 parts commission. 

(2) Hence ^ of 90 ct. = $^0, 2d commission. 

(3) ji of Hro of 90 ct. = $sV loss by the decline. 



70 FAIRCHILD'S SOLUTION BOOK. 

(4) f§-£ of %%% of ±%% of 90 ct. = 10 ct, commission on hogs 

from selling them. 

(5) In all my loss on one dollar of the receipts for cattle = 

10 ct. + $^ + 1^+10 ct. = $ T \. 

(6) Now, my loss on the $18 is first ^\ of $18 = $f|f, and 

CK 2 

secondly -^ of \U of $18 = $ff§-; and thirdly ifg of 

m ° f m of $ 18 = $ 2 - 

(7) ••• $f -f* + UM + §2 = *6, loss on $18. 

(8) $86 — $6 = $80, loss on cattle receipts. 

(9) If the loss on $1 of the receipts for cattle is $^, then, 

there must have been as many dollars in the receipts 
for cattle as T \ is contained times in $80, or $200. 

.•'. $200 = the sale of the cattle. 

Note. — It will be noticed that we keep separate the $1 received for 
cattle, and the $18 increase. 

PROBLEM 153. 

An agent received $4325 to invest in mess pork at $16 per barrel, 
after deducting- his purchasing commission of 4%: if the charges for 
incidentals were $81.40, besides cartage of 75 ct. per load of 8 barrels, 
how many barrels did he buy, and what unexpended balance does he 
place to the credit of his principal? 

Solution. 

( 1) The commission on one barrel of pork is $16 X .04 = 64 ct., 

and the cartage $.75 -f 8 = $.09f, making a cost of $16 
+ $.64 + $.09f = $16.73f per barrel, besidesincidentals. 

(2) $4325 — $81.40 -r $16.73f = 253, number of barrels. 

(3) $16 X 253 = $4048, the cost of the pork. 

(4) $4048 X .04 = $161.92, commission. 

(5) 253 -7- 8 = 31|, number of loads carted. 

(6) 3 If X 75 ct. = $23.71f, cost for cartage. 

(7) Total expenditure = $161.92 + $81.40 + $23.71| + $4048 

= $4315.03f 

(8) $4325 — $4315.03| = $9.96|, unexpended balance. 

.*. $9.96^ is the unexpended balance. 

PROBLEM 154. 

I received from Hyson & Son, of Chicago, a ship load of corn, 
which I sold for 60 ct. per bushel, on a commission of 4%; and by the 
shipper's instructions invested the net proceeds in barley at T5 ct. per 
bu., charging 5% for buying; my total commission was $1350: how 



COMMISSION. 71 

many bushels, of corn did Hyson & Son ship and how many bushels of 
barley should they receive? 

Solution. 

(1) Out of every $1 of receipts for corn I received 4 ct., and 

had 96 ct. proceeds to invest in barley. Now, this 96 
ct. is made up of 100 parts and 5 parts commission. 

(2) Then, T fe of 96 ct. = f§$ ct., 2d com. 

(3) 4 ct. + fff ct. = m ct., or $1^, whole com., which is 

$1350. 

(4) Then, there must have been as many dollars in the re- 

ceipts for corn as $jfe is contained times in $1350, or 
$15750; and there were 15750 -^ .60 = 26250, bu. of 
corn; ($15750 — $1350) -^ .75 = 19200, bu. of barley. 

.". 26250 = no. bu. of corn, and 19200 = no. bu. of barley. 

PROBLEM 155. 

A Gilboa brewer remitted $21500 to a Leipsic commission merchant 
with instructions to invest 40% of it in barley, and the remainder, less 
all charges, in hops; the agent paid 60 ct. per bushel for barley, and 
20 ct. per pound for hops, charging 2% for buying the barley, 3% for 
buying the hops, and 5% for guaranteeing the quality of each purchase: 
if his incidental charges were $187.50, what quantity of each product 
did he buy, and what was the amount of his commission? 

Solution. 

(1) 40% of $21500 = $8600, cost of the barley. 

(2) $8600 -r- .60 = 14333, no. of bu. of barley. 

(3) $21500 — $8600 = $12900, what he had left. 

(4) 2% of $8600 = $172, com. for buying barley. 

(5) 5% of $8600 = $430, for guaranteeing the purchase. 

(6) $8600 + $172 + $430 + $187.50 = $9389.50. 

(7) $21500 — $9389.50 = $12110.50. 

(8) The hops cost $12110.50 ~ 1.08 = $11213.42*f 

(9) Com., 3% of $11213.42 = $336.40. 

(10) The hops weighed $11213.42£f -r- .20 = 56067^ lb. 

(11) Total commission is $336.40 + $172 = $508.40. 

.'. He bought 14333 bu. of barley and 56067-& lb. hops, 
for a commission of $508.40. 

PROBLEM 156. 

I ordered an agent in Bluffton to buy flour, which I afterwards sold 
at 20% profit, and gained $1.56 per barrel: if his rate of commission 
was 4%, and his total commission $23.40, how many barrels did he buy? 



72 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) Out of every $1 invested in flour the agent received 4 ct., 

and $23.40 = amount the agent received. 

(2) Then, there must have been as many dollars invested in 

flour as 4 ct. is contained times in $23.40, or $585. 

(3) I also made 20 ct. profit by selling it; I made $1.56 per 

barrel. Then, 1 barrel of flour must have cost as much 
as 20 ct. is contained times in $1.56, or $7.80. 

(4) Hence, $585 + $7.80 = 75 bbl. 

.*. 75 bbl. were bought. 

PROBLEM 157. 

I sold at a commission of 4%, and reinvested at 20% ; if I had sold 
at 20% and reinvested at 4%, my commission would have been $153.85 
more: "what was the amount of sales? 

Solution. 

(1) Out of every $1 receipts for the article sold I received 4 

ct. and had 96 ct. to reinvest. Now, this 96 ct. is made 
up of 100 parts investment and 20 parts commission. 

(2) Then ^o of 96 ct - = W c t-> my 2d com. 

(3) .*. Out of every $1 received for the article, I get 1st 4 ct. 

and 2d ^o 1 ct -» or 20 ct. commission by 1st condition. 

(4) Out of every $1 receipts for the article sold I receive 20 

ct. and have 80 ct. to reinvest. Now, this 80 ct. is made 
up of 100 parts investment and 4 parts commission. 

(5) Then yfo of 80 ct. = f^J- ct., my 2d com. 

(6) .'. Out of every $1 received for the article, I get 1st 20 ct. 

and 2d f|-J ct., or 23-^ ct. com. by the 2d condition. 

(7) The difference is 23 T V ct. — 20 ct., or 3 T V ct., i. e. y I would 

receive 3^ ct. more by 2d condition; I also would re- 
ceive $153.85 more. 

(8) .'. The amount of sales was as much as d-^ ct. is contained 

times in $153.85, or $5000+. 

.*. $5000+ = the amount of sales. 

PROBLEM 158. 

A sold a consignment of cotton, losing- 4% ; keeping- $18 of the pro- 
ceeds, he gave the remainder to an ag-ent to buy sug-ar, 8% commission; 
he lost in all $32: find the value of the cotton. 

First Solution. 

(1) Let $1 = the value of the cotton. 

(2) Then, 96 ct. = proceeds. 



COMMISSION. 73 

(3) Now, this 96 ct. is made up of 100 parts investment and 

8 parts commission. 

(4) /. The investment was {%$ of 96 ct., or 88f ct. 

(5) Had no money been reserved, the loss would have been 

11^ ct. on each $1 of the first value; or, in other words, 
he would have \ of the whole had he left $18 with the 
proceeds. But the loss on that $18 would have been 
the commission, y-g-g- of $18 = %\\. 

(6) If he had kept no money, the whole loss would have been 

$32 + %1\ — $33J; and this would have been \ of the 
value of the cotton; that must have been 9 X $33^ = 
$300. 

Second Solution. 

(1) Out of each $1 received for cotton, my agent received 4 

ct. commission. 

(2) $1 — 4 ct. = 96 ct., proceeds of sale. 

(3) Now, this 96 ct. is made up of 100 parts investment and 8 

parts commission. 108 = whole no. of parts. 

(4) yfj of 96 ct. = 7 SI ct., com. 

(5) T fg- of $18 = $lf|, com. on $18, providing he had not re- 

served it. 

(6) 4 ct. + f§| ct. — $|4f = whole loss. 

(7) .'. 432 ct. + 768 ct. — $144 = $3450, or 

(8) 1200 ct. = $3600; then there will be as many dollars in- 

vested in cotton as 1200 ct. is contained times in $3600, 
or $300. 

.'. $300 was invested in cotton. 



CHAPTER VIII. 



STOCKS AND BONDS. 



48. Stock is capital invested in business. 

49. A Bond is a written or printed obligation, under seal, 
securing the payment of a certain sum of money at or before 
a specified time. 

50. A Dividend is a sum of money to be paid to the stock- 
holders in proportion to their amounts of stock. 

51. Par Value of money, stock, drafts, etc., is the nominal 
value on their face. 

52. Market Value is the value for which they sell. 

53. Discount is the excess of the par value over the market 
value. 

54. Premium is the excess of the market value over the par 
value. 

55. Brokerage is the sum paid the agent for buying stock. 

56. Buying Lojig. — Buying in the expectation of a rise. 
Kite Flying is expanding one's credit beyond wholesome 

limits. 

Short Selling is selling for future delivery what one does not 
have, in hopes that prices will fall. 

Ballooning is working up a stock far beyond its intrinsic 
worth by favorable stories or fictitious sales. 

Watering Stock is the art of doubling the quantity of stock 
without improving its quality. 

Forcing Quotations is where brokers wish to keep up the price 
of stock and to prevent its falling out of sight. This is gen- 
erally accomplished by a small sale. 
(74) 



STOCKS AND BONDS. 75 

Corner. — The buying up of a large quantity of stock or 
grain to raise the price, when the market is oversold; the 
shorts, if compelled to deliver, find themselves in a "corner." 

Pool. — The stock or money contributed by a clique to carry 
through a corner. 

Pointer. — A theory or fact regarding the market on which 
one bases a speculation. 

Gunning a Stock is using every art to produce a break when 
it is known that a certain house is heavily supplied and would 
be unable to resist an attack. 

PROBLEM 159. 
What is the market value of 300 shares of R. R. stock at 80? 

Solution. 

(1) 300 X $100 = $30000, face value of stock certificate. 

(2) 80 ct. = the cost of $1 of this certificate. 

(3) $30000 X 80 ct. = $24000. 

.-. The certificate cost $24000. 

PROBLEM 1(50. 

If a man invests $6320 in 6% bonds, at 80, what will be his annual 
income from the investment? 

Solution. 

(1) $6320 = cost of the bonds. 

(2) 80 ct. = cost of $1 of the bonds. 

(3) As many dollars of these bonds can be bought as 80 ct. is 

contained times in $6320, or $7900, face value of bonds. 

(4) 6 ct. = income on $1 of bonds; then $7900 X .06, or $474, 

is the income. 

.'. $474 = total income on the bonds. 

PROBLEM 161. 

When g-old was at a premium of 30%, what was $1000 in gold worth 
in currency? 

Solution. 

(1) $1 in gold = $1.30 in currency. 

(2) $1000 in gold = $1000 X 1.30, or $1300 in currency. 

PROBLEM 162. 
When gold was at 112, what was the value in gold of SI in currency? 



76 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) 112 ct. in currency = $1, or 100 ct. in gold. 

(2) 1 ct. in currency = T {^ of 100 ct., or f J ct. in gold. 

(3) $1 in gold = 100 X ff ct., or 89f ct. in gold. 

.'. $1 in currency, by the given data, would be worth 89f 
ct. in gold. 

PROBLEM 163? 

A invested $28000 in Lake Shore Railroad stock, at 70 %. If the 
stock yields 8% annually, what is the amount of his income. 

(R. H. A.,p. 222.) 
Solution. 

(1) $2800 = cost of railroad stock; 70 ct. = cost of $1 of the 

stock. 

(2) 128000 -rr 70 ct. = $40000, face value of railroad stock. 

(3) 8 ct. = income on $1 of this stock, and $40000 X .08 = 

$3200, income received. 

PROBLEM 164. 

Which is the more profitable, to invest $10000 in 6% stock purchased 
at 75%, or in 5% stock purchased at 60%, allowing- brokerage £%? 

{R. H, A., prod. 6, p. 223.) 
Solution. 

(1) $10000 = cost of 6% stock; 75 ct. + \ ct. = 75| ct., cost 

of $1 of this stock. 

(2) Then, $10000 -r- 75^ ct. = $13245.03, face value of the 6% 

stock. 

(3) 6 ct. = income of $1 of this stock, and $13245.03 X 6 ct. 

= $794.7018, income received on the 6% stock. 

(4) $10000 = cost of 5% stock; 60 ct. + | ct. = 60^ ct., cost 

of $1 of this stock. 

(5) Then, $10000 -J- 60fc ct. = $16328.90, face value of the 5% 

stock. 

(6) 5 ct. — income of $1 of this stock, and $16528.90 X 5 ct. 

= $826.44, income received on the 5% stock. 

.-. $826.44 — $794.70 = $31.74 in favor of the 5% stock. 

PROBLEM 165. 

A has a farm valued at $46000, which pays him 5% on the invest- 
ment; through a broker, who charges $56.50 for his services, he ex- 
changes it for insurance stock at 9% premium, and this increases his 
annual income by $1090: what dividend does the stock pay? 

(R. H. A.,p. 222.) 



STOCKS AND BONDS. 77 

Solution. 

(1) $46000 = the value of the farm; 5 ct. = income on every 

$1 invested in the farm. 

(2) $46000 X 5 ct. = $2300, income on the farm. 

(3) 109 ct. = cost of $1 of insurance stock; then ($46000 — 

56.50) -f- 109 ct. = $42150, the par value of the insur- 
ance stock. 

(4) $2300 + $1072 = $3372, income on the insurance stock. 

(5) $3372 -f- $42150 = 8 ct., or 8%. 
.*. The stock pays 8% dividend. 

PROBLEM ]66. 
Howard has at order $122400, and can allow brokerage \ c /c , and buy 
insurance stock at 101£%, yielding 4£%; but if he send to the broker 
$100 more for investment, and buy rolling-mill stock at 103£%, the in- 
come will only be half so large: what rate does the higher stock pay? 

{R. H. A.,p. 224.) 
Solution. 

(1) $122400 = amount of order; each $1 of the insurance 

stock cost 101-| ct. + 1 ct -» or 10- ct -> an d yields 4j ct. 

(2) $122400 -r $1.02 = $120000, the par value of the insurance 

stock. 

(3) $120000 X 4i ct. = $5000, total income on the insurance 

stock. 

(4) $122500 = amount invested in rolling-mill stock. 

(5) Each $1 of the rolling-mill stock cost 103£ ct. -f \ ct., or 

104 ct.; then $122500 -7- 104 ct. = $- 1 | 2 V°, the par val- 
ue of the rolling-mill stock. 

(6) The income being $2500, the rate = $2500 -~ ^|^p = 

i^t,or2A%. 
.'. The rolling-mill stock pays 2^%. 



PROBLEM 167. 
What amount is invested by A, whose canal stock, yielding 1% , 
brings an income of $300, but sells in market for 92%? 

Solution. 

(1) Each $1 of the canal stock yields 4 ct.; $300 = whole in- 

come; then $300 -r .04 = $7500. 

(2) 92 ct. = cost of $1 of this stock; then 7500 X 92 ct. = 

$69000, amount invested. 

PROBLEM 168. 
If I invest all my money in 5% furnace stock, salable at 75%, my 
income will be $180: how much must I borrow to make an investment 



78 FAIRCHILD'S SOLUTION BOOK. 

in 6% state stock, selling- at 102% , to have that income? 

{R. H. A., p. 225,prob. 2.) 
Solution. 

(1) Each $1 invested in the furnace stock yields 5 ct. 

(2) $180 = whole income on the furnace stock; then $180 — 

.05 = $3600, par value of furnace stock. 

(3) 75 ct. = cost of $1 of this stock; then $3600 X .75 = 

$2700, whole cost of furnace stock, and this is all my 
money. 

(4) Each $1 invested in state stock yields 6 ct.; $180 = 

whole income on the state stock; then $180 -=- .06 = 
$3000, par value of state stock. 

(5) 102 ct. = cost of $1 of this stock; then $3000 X 1.02 = 

$3060, cost of state stock. 

(6) I have only $2700; I must borrow $3060 — $2700 = $360. 

PROBLEM 169. 

If railroad stock be yielding- 6%, and is 20% below par, how much 
would have to be invested to bring- an income of $390? 

[R.H.A.,p.22 5 .) 
Solution. 

(1) $390 = income desired; 6 ct. = income of $1 of railroad 

stock. 

(2) $390 -f- 6 ct. = $6500, par value. 

(3) 80 ct. = cost of $1 of this stock; then $6500 X .80 = 

$5200, cost of the railroad stock, or amount invested. 

x PROBLEM 170. 

If 6% pike stock is worth 40% less than 8% g-as stock, and the in- 
come on each be $900, what money is invested in each, if investment 
pays 5%? 

Solution. 

(1) 6 ct. = income on $1 of pike stock; $900 = income on 

pike stock. 

(2) $900 -=- .06 = $15000, value of pike stock. 

(3) 8 ct. = income on $1 of pike stock; $900 = income on 

gas stock; then, $900 -r- .08 = $11250, value of gas 
stock. 

(4) If pike stock is 40%, or |- less in market than gas stock, 

then the cost of $1 of pike stock would be f of the cost 
of $1 of gas stock, but we see by comparison of the two 
values that there are f as many dollars of pike stock as 
there are in gas stock. | of -J of gas stock = -J-J of gas 
stock = cost of pike stock in terms of gas stock. 



STOCKS AND BONDS. 79 

(5) if = cost of gas stock, and y| + t! — ff» tota l invest- 

ment in both. 

(6) $900, income on pike stock + $900, income on gas stock 

=$1800, total income. 

(7) Every $1 of investment yielded 5 ct. .'. $1800^- .05 = 

$3600, total investment. 

(8) But fj = total investment; .'. f£ = $36000. 

(9) T ^ = J T of $36000 = $13331. 

(10) || = 12 times $1333^ = $16000, cost of pike stock. 

(11) if = 15 times $1333J = $20000, cost of gas stock. 

-PROBLEM 171. 

A's whole income on 3J-% stock at 6|% discount, and 2£-% bonds at 
12a % premium, is $9000: what was invested in each, if the income on 
the latter was 25% more than on the former? 

Solution. 

(1) Every $1 invested in stock cost 93J ct., and yields 3 J- ct. 

(2) The income is 3J -f- 93J, or -£% of the investment. 

(3) Every $1 invested in bonds cost 112-J- ct., and yields 2-J- 

ct.; .'. the income is 2-J- -=- 112-J-, or -fe of the investm't. 

(4) Since the latter income was 25%, or \ more than the 

former, then the former income is J and the latter -J. 

(5) £ + -£ = f, the whole income. 

(6) | = $9000. 

(7) I = i of $9000 = $1000. 

(8) f = 4 times $1000 = $4000, income of former. 

(9) i = 5 times $1000 = $5000, income of latter. 

(10) ft = $4000. 

(11) || = 28 times $4000 = $112000, investment in stock. 

(12) ft = $5000. 

(13) || = 45 times $5000 = $225000, investment in bonds. 

^PROBLEM 172. 

I invest $3600 more in 7-i% bank stock at 12£-% premium, than in 
8% canal stock at 4% discount; from the latter my income is 25% less 
than from the former: what is the face value of each stock and my 
whole income? 

Solution. 

(1) Every $1 invested in bank stock costs 112| ct. and 

yields !\ ct. .'. the income is 1\ ~ 112.V, or ft of the 
investment. 

(2) Every $1 invested in canal stock costs 96 ct., and yields 

8 ct.; .*. the income is ^ 8 g-, or y 1 ^ 



80 FAIRCHILD'S SOLUTION BOOK. 

(3) Since the income from the latter is 25%, or \ less than 

the former, then the income from the latter is f of the 
income from the former; J of ^ the former income = 
^% of the former. 

(4) T *£ of the latter investment — g% of the former. 

(5) If of the latter = f§ of the former. 

(6) |4 = former inv. — |-§- = f-f, diff. in investments. 

(7) ••• U = 13600. 

(8) gV = -h of P600 = |150. 

(9) fj = 60 times $150 = $9000, inv. in bank stock. 

(10) Also 36 times $150 = $5400, inv. in canal stock. 

(11) T V of $9000 = $600, income from bank stock. 

(12) jL of $5400 = $450, income from canal stock. 

(13) The face value of the bank stock is $9000 ~ 1.12$ = 

$8000, and that of canal stock is $5400 -r- .96 = $5625. 

PROBLEM 173. 

For what must one buy a 5% stock to realize an income of 6% on 
his investment? 

Solution. 

(1) 5 ct. = income of $1 of the stock; the income is to be 6% 

of the investment; this 5 ct. must be 6% of the cost of 
$1 of stock. 

(2) 6% of the cost of $1 of stock = 5 ct. 

(3) 1% of the cost of $1 of stock = $ of 5 ct. or f ct. 

(4) 100% = 100 times £ ct. = 83$ ct., cost of $1 of stock. 

PROBLEM 174. 

J. W. Burris, through his broker, buys N. Y. 6's at 107-£-, and twice 
as much in U. S. 5's of '82 at 98£, brokerage in each case \% ; his an- 
nual income from both is $3348: how much was paid for each kind of 
stock? 

Solution. 

(1) 107$ ct. -f- $ ct. = 108 ct., cost of $1 of N. Y. 6's. 

(2) $1 invested will buy $$-$£ of N. Y. stock, yielding $^, 

or fcfa per annum. 

(3) 98$ ct. + $ ct. = 99 ct., cost of $1 of U. S. stock. 

(4) Then, $1 invested will buy $±£?- U. S. 5's, and will yield 

$q 5 q per annum. 

(5) $2 invested will yield 2 times $^, or $$£. 

(6) $& + #$§- = #;ftV, income on $3 of stock. 

(7) $ of $ T Vg = $■££%, average income on $1 of the whole 

stock. 



STOCKS AND BONDS. 81 

(8) $3348, entire income -f- -gfe = $64152, whole am't inv. 

(9) For every $1 invested in N. Y. 6's, $2 are invested in U. 

S. 5's; .-. i of $64152 = $21384, amount in N. Y. 6's. 
(10) |- of $64152 = $42768, amount in U. S. 5's. 

PROBLEM 175. 

If 6% express stock is worth in market 20% more than 10% railroad 
stock, and the railroad stock 25% more than 12% telegraph stock, then 
at what rate was each bought and what was each investment, if the in- 
come from each is $1200 and total investment pays 10[-$-%? 

Solution. 

(1) 6 ct. = income on $1 of express stock; $1200 = whole 

income on express stock; then, $1200 -f- .06, or $20000 
= entire value of express stock. 

(2) 10 ct. = income on $1 of railroad stock; $1200 = entire 

income; then, $1200 -f- .10, or $12000 = face value of 
railroad stock. 

(3) 12 ct. = income on $1 of telegraph stock; $1200 = en- 

tire income; then, $1200 -f- .12, or $10000 = face value 
of telegraph stock. 

(4) If the railroad stock is 25%, or \, better than telegraph 

stock, then the cost of $1 of railroad stock would be J 
of the cost of $1 of telegraph stock. 

(5) We see by comparison of the two face values that there 

are f as many dollars of railroad stock as there are of 
telegraph stock; .'. J of |, or J of telegraph stock = 
cost of railroad stock in terms of telegraph stock. 

(6) | = cost of telegraph stock. 

(7) Now, if express stock is 20%, or i, better than railroad 

stock, then the cost of $1 of express stock would be | 
of the cost of railroad stock; i e., if the railroad stock 
were the basis of comparison; but we take the tele- 
graph stock. 

(8) By comparing the express stock and the telegraph stock 

there are j as many dollars of express stock as there 
are of telegraph stock; .". J of \ = !} of telegraph stock 
= cost of express stock in terms of telegraph stock. 

(9) -f + f + I — V» or entire cost of the stocks. 

(10) $1200 = the income of each stock; hence, $1200 X 3, or 

$3600 = total income. 

(11) Each $1 invested yielded an income of Y0\\ ct.; hence, 

$3600 -r- 10{? ct. = $33000, entire cost of investment; 
also, y == entire cost of investment. 

(12) V = $33000. 



82 FAIR CHILD'S SOL U TION BOOK. 

(13) \ = T V of 133000 = $3000. 

(14) | = 2 times $3000 = $6000, cost of telegraph stock. 

(15) | = 3 times $3000 = $9000, cost of railroad stock. 

(16) f = 6 times $3000 = $18000, cost of express stock. 

(17) The express stock was bought at $18000 -f- $20000 = 90%. 

(18) The railroad was bought at $9000 -i- $12000 = 75%. 

(19) The telegraph was bought at $6000 -r- $10000 == 60%. 

PROBLEM 176. 

Invested in U. S. 4-|-'s at 105, brokerage | % ; f as much in U. P. 6's 
at 119f , brokerage \% ; and 3 times as much in N. Y. 7's at 87^, broker- 
age \%\ if my entire income is $1702, find my investment. 

Solution. 

(1) 105 ct. + i ct. = 105^ ct, total cost of $1 of 1st. 

(2) Then, $1 invested will buy T ^ of U. S, 4^'s, and will 

yield $ T ^ 5 ^, or $2tt P er annum. 

(3) 119| ct. + \ ct. = 120 ct., total cost of $1 of 2d. 

(4) Then, $1 invested will buy %\\% of U. P. 6's, and will 

yield $y-f-o per annum. 

(5) $f invested will yield £ of %&^ or %fc. 

(6) 87^ ct. + \ ct. = 87| ct., cost of $1 of the 3d. 

(7) Then, $1 invested will buy $££$ of N. Y. 7's, and will 

yield $-g-if 5-, or $ T 1 T 4 3 - per annum. 

(8) $3 invested will yield 3 times $^, or %fe. 

(9) $2TT + $i + $A = UiH> income on $4f of stock. 

1 * flsQ'T'^ 7 

(10) -r- of $5^7!- — -K^p, av'ge income on $1 of entire stock. 

(11) $1702, the total income H- ^glffi , the income on $1 = 

$25320, total amount invested. 

(12) For every $1 invested in U. S. 4^'s, $-| are invested in U. 

P. 6's, and $3 invested in N. Y. 7's. 

(13) .-. -^ of $25320 = $5275, amount invested in U. S. 4fs. 

(14) I of $5275 = $4220, amount invested in U. P. 6's. 

(15) $5275 X 3 = $15825, amount invested in N. Y. 7's, 

(16) .-. $25320 = the whole investment. 

PROBLEM 177. 

Suppose I sell $15000 in U. S. bonds at 102£, and invest part of the 
proceeds in 7-J- state stock at 106, realizing an income of $720, brokerage 
in each case \%, and buy a house with the remainder: at what must I 
rent it per month to make 5% interest on my investment? 



STOCKS AND BONDS. 83 

Solution. 

(1) $1.02J — i ct. = $1.02, amount received from the sale of 

$1 of stock. 

(2) $15000 X 1.02 = $15300, received from the sale of the 

bonds. 

(3) $720 = income on state stock, and 1\ ct. = income on $1 

of state stock. 

(4) $720 -=- .075 = $9600, face value of state stock. 

(5) $9600 X 1.06| = $10200, the cost of the state stock. 

(6) Hence, the house cost $15300 — $10200, or $5100. 

(7) 5 ct. = income on $1 for 1 year's rent. 

(8) $5100 X .05 = $255, received for 1 year's rent. 

(9) $255 -7- 12 = $21.25, received per month. 

PROBLEM 178. 

A paid $1075 for U. S. 5-20 6% bonds at \\% premium, interest pay- 
able in g"old semi-annually: when the average premium on gold was 
112%, did he make more or less than B, who invested an equal sum in 
railroad stock at 14% below par, which paid a semi-annual dividend of 
4%? 

Solution. 

(1) $1071 = cost of $1 of U. S. 5-20's. 

(2) $1075 -r- 1.075 = $1000, face value of 5-20's, and $30 is 

the interest in gold, or $33.60 in currency. 

(3) 86 ct. = cost of $1 of railroad stock. 

(4) $1075 -r- .86 = $1250, face value of railroad stock, and 4 

% of $1250 = $50, interest on the stock. 

(5) Hence, $50 — $33 60 = $16.40, what A makes less than 

B every 6 months. 

PROBLEM 179. 

I invested $2700 in stock at 25% discount, which pays 8% annual 
dividends: how much must I invest in stock at 4% discount and pay- 
ing- 10% annual dividends, to secure an equal income? 

Solution. 

(1) $2700 = cost of the 1st stock; 75 ct. = cost of $1 of this 

stock. 

(2) $2700 -r- .75, or $3600 = face value. 

(3) 8 ct. =± income on $1 of this stock; $3600 X .08, or $288 

= income realized. 

(4) 10 ct. = income on the 2d stock. We see that the in- 

comes are equal; then — 

(5) $288 -T- .10 = $2880, the par value of the stock. 



84 FAIRCHILD'S SOLUTION BOOK. 

(6) 96 ct. = the cost of $1 of this stock; then, $2880 X .96 = 
$2764.80, cost of the stock, or the investment. 

PROBLEM 180. 

I invest a certain sum in 6's at 85, and the same sum in 7's at 95, 
receiving - $5 a year more from the latter investment: how much do I 
invest in each? 

Solution. 

(1) 85 ct. = cost of $1 of the 1st; then, $1 invested will buy 

$ J 8 Q 5°- of 1st stock, and yield $ F 6 F per annum. 

(2) 95 ct. = cost of $1 of 2d stock; then, $1 invested will buy 

%Hi>- °f ^d stock, and yield %fe per annum. 

(3) $ ¥ 7 -5 — $-§-%- = $yjf 5-, difference in incomes of the two 

stocks. 

(4) $5 = difference in investments; then, $5 -7- $ 8 of 5 — 

$1615, what I invest in each. 

(5) g 7 ^ of $1615 = $119, income of 2d stock. 

(6) ¥ 6 g of $1615 = $114, income of 1st stock. 

PROBLEM 181. 

A man bought Michigan Central at 120, and sold at 124: what % 
of the investment did he gain? 

So hit ion. 

(1) $1.20 = cost of $1 of this stock, and $1.24 = the selling 

price; $1.24 — $1.20 = 4ct., gain. 

(2) Then, 4 ct. -r- 81.20 = 3J%', gain on the investment. 

PROBLEM 182. 

I bought U. S. 4% bonds at 119£%, brokerage \% additional, and 
derive from the purchase a quarterly income of $300: how much did I 
invest? 

Solution. 

(1) $300 = income for 3 months, and $300 X 4 = $1200, in- 

come for 1 year. 

(2) 4 ct. = income of $1 of these bonds; then, $1200 -r- .04 = 

$30000, face value. 

(3) 119| + 1 :•■= $1.20, cost of $1 of this stock. 

(4) $30000 X $1.20 = $36000, the cost or investment. 

PROBLEM 183. 

What is my gain or loss, if I buy 112 shares of stock in a transpor- 
tation company, at life premium, and, after receiving a dividend of 
9% , sell it for 8% less than it cost me? 



STOCKS AND BONDS. 85 

Solution. 

(1) Let $100 represent the cost of each share. 

(2) $100 X 112 = $11200, face value of the stock. 

(3) $1.17 = the cost of each $1 of this stock. 

(4) $11200 X $1.17 = $13104, the cost of the stock. 

(5) $11200 X .09 = $1008, the dividend, and 92 ct., the sell- 

ing price of this stock. 

(6) Then, the amount realized from sale is $13104 X .92 = 

$12055.68. 

(7) .-. I sustained a loss of $13104 — ($12055.68 + $1008) = 

140.32. 

PROBLEM 184. 
I hold 4% stock and at the end of the year I take my dividend in 
same stock at 20% discount; the par value of my stock is then $10500: 
find dividend. 

Solution. 

(1) 4 ct. = income on $1 of the stock, which dividend I take 

in the same stock. 

(2) I get $ 8 \ of the stock as dividend. 

(3) $1 + $g%, or $|| — par value of $1 of the stock. 

(4) .-. $10500 -r $|J = $10000, the original value. 

(5) fa of $10000 = $500, dividend of face value of the stock. 

PROBLEM 185. 
Bought stock at 20% discount and sold to gain 12£% on my invest- 
ment: if I invest proceeds again, at what discount must I buy to yield 
at next selling 35% on first investment? 

Solution. 

(1) Let $100 represent the value of the stock; 80 ct. = the 

cost of $1 of this stock. 

(2) $100 X 80 ct. = $80, the cost of the stock. 

(8) 81.124 = the selling price of 81 of the stock; $80 X $1.12| 

= $90, selling price. 

(4) ($100 — $90) -r $100 = 10%, the discount at which it is 

sold. 

(5) But at the second sale I am to make 35% profit on first 

investment, I must sell for $80 X 1.35 = $108. 

(6) As the stock was sold at 10% discount, the face must be 

$108 -f- .90 = $120. 

(7) Then ($120 — $108) + 120 = 10%, discount required. 

PROBLEM 186. 
How many shares of railroad stock ($50 each) at 541, must be sold 
in order that the proceeds, invested in 6's at 95, may yield an income of 
$750, not considering- brokerage? 



CHAPTER IX. 



INSURANCE. 



57. Insurance is an indemnity against loss. It is of two 
kinds: Property Insurance and Personal Insurance. 

58. Property Insurance is security against loss by fire or 
transportation. Insuring anything is called "taking a risk." 

59. Fire Insurance is security against loss by fire. 

60. Marine Insurance is security against loss by navigation. 

61. The Policy is the written agreement or contract between 
the insurers and the insured. 

62. Premium is the sum charged for insurance; it is a cer- 
tain rate per cent of the amount insured. 

PROBLEM 187. 

Insured a house for $2500, and furniture for $600, at T %% : what was 
the premium? {/?. H. A., p. 230.) 

Solution. 

(1) 100% — am't insured. 

(2) $2500 + $600 = $3100. 

(3) 100% = $3100. 

(4) 1% = T £o of $3100, or $31. 

(5) -&% = T <v times $31 = $18.60, premium. 

.". $18.60 is the premium. 

PROBLEM 188. 

I insured property at 2% ; reinsured $8000 at If %, and $10000 of it at 
2i% : what was the amount, my share of the premium being- $207.50? 

Solution, 

(1) 100% = amount insured; 2% = premium. $8000 = 

amount reinsured 1st time. 

(2) If % of $8000 = $140, premium on $8000. 

(86) 



INSURANCE. 87 

(3) 1|% of $10000 = $212.56, premium on $10000. 

(4) $207.50 = my share of the premium; then, $140 + $212.50 

+ $207.50 = $560, whole premium. 

(5) 2% = premium = $560. 

(6) 1% = \ of $560 = $280. 

(7) 100% = 100 times $280 = $28000. 

.-. $28000 = amount insured. 

PROBLEM 189. 

I paid $180 for insuring my stock for f of its value at 3% : what is 
the value of the stock? 

Solution. 

(1) 100% = value of J of the stock. 

(2) 3% = premium = $180. 

(3) 1% = \ of $180, or $60. 

(4) 100% = 100 times $60 = $6000, value of § of the stock. 

(5) | of the stock = $6000. 

(6) i = -J of $6000, or $3000. 

(7) f the stock = 3 times $3000 = $9000. 

.-. $9000 = the value of the stock. 

PROBLEM 190. 

I insured, a house at \\% ; reinsured § of it at 2% , and \ of it at 2i ■% : 
what rate of insurance do I get on the remainder? (J?. H. A., p. 231.) 

Solution. 

(1) 100% = risk; 1|% = premium. 

(2) f of 100% = 40%, amount reinsured at 2%. 

(3) 2% of 40% = .80%, amount paid out for reinsuring |- of 

the risk. 

(4) I of 100% = 25%, amount reinsured at 2|%. 

(5) 2\°/ of 25% = .62|%, amount paid out for reinsuring \ of 

the risk. 

(6) .80% + .62^ = 1.42-|%, amount of premiums paid out. 

(7) 1\% — 1.42-|% = .07^%, amount of premium left. 

(8) 40% + 25% = 65%, amount reinsured. 

(9) 100% — 65% = 35%, risk left. 

(10) 35% = 100% of itself. 

(11) 1% = & of 100% = -yv°%. 

(12) .07^% = .07| times ^ = ^ %, rate of premium. 
• "• i : V% = rate of insurance that I receive. 



88 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 191. 

Took a risk at 2% : reinsured $10000 of it at 2£% , and $8000 at If % ; 
my share of the premium was $207.50: what sum was insured? 

{R.H.A.,p.2 3 i.) 
Solution. 

(1) 2|% of $10000 = $212.50, amount paid out on $10000. 

(2) If % of $8000 = $140, amount paid out on $8000. 

(3) $212.50 + $140 = $352.50, whole amount paid out. 

(4) $207.50 = amount I receive. 

(5) $352.50 + $207.50 = $560, premium on whole risk. 

(6) 100% = risk. 

(7) 2% = premium = $560. 

(8) 1% = 1 of $560 = $280. 

(9) 100% = 100 times $280 = $28000, risk. 

.*. $28000 = what was insured. 

PROBLEM 192. 

The Mutual Fire Insurance Company insured a building- and its 
stock for | of its value, charging- If % . The Union Insurance Company 
relieved them of \ of the risk, at \\%. The building- and stock being- 
destroyed by fire, the Union lost $49000 less than the Mutual: what 
amount of money did the owners of the building- and stock lose? 

{R. H. A., p. 232.) 
Solution. 

(1) 100% = value of property insured. 

(2) f of 100% = 66f, the whole risk. 

(3) 1|% of 66f% = 1|%, entire premium. 

(4) \ of 66|% = 16|%, Union's risk. 

(5) 66|% — 16f % = 50%, Mutual's risk. 

(6) \\°] of 16f % =1%, Union's premium. 

(7) l-i-% — i% = H%, Mutual's premium. 

(8) 50% — }i% = 49 t l loss of Mutual. 

(9) 16f % — i% = 16^%, loss of Union. 

(10) 49 T !-2% — 16^% = 32f %, excess of Mutual's loss. 

(11) 32f % = $49000. 

(12) 1% = $49000 -f- 32| = $1500. 

(13) 100% — 66f % + H% = 34|% =owners' loss. 

(14) 34^% = 34| times $1500 = $51750. 

.*. $51750 = owners' loss. 



CHAPTER X. 



I. INTEREST. 



63. Interest is money paid for the use of money. 

64. The Principal is the sum on which interest is charged. 

65. The Rate of interest is the rate per cent on $1 for one 
year. 

66. The Time is the period during which the money is on 
interest. 

67. The Amount is the sum of the principal and the interest. 

68. Simple Interest is interest on the principal only. 

69. Compound Interest is interest on the principal and in- 
terest. 

70. Legal Interest is interest at the rate fixed by law. 

71. Usury is a rate of interest greater than the law allows. 

Note. — In notes, contracts, mortgages, etc., when no rate is speci- 
fied, the legal rate is understood, 

72. The Six Per Cent Method is so called because the pro- 
cess is based upon that rate. 

Rule. — Multiply the member of years by the rate, take \ of the 
number of months as cents, and \ of the number of days as mills ; 
their sum will be the interest of $1 for the given time at 6%. 

PROBLEM 193. 
Required the interest of $380 for 3 yr. 4 mo. 12 da., at 6%. 

Solution. 

(1) The interest on $1 for 1 yr. is 6 ct., and for 3 yr. it is 8 

times 6 ct., or $.18. 

(2) For 4 mo. it is \ of 4, or $.02. 

(89) 



90 FAIRCHILD'S SOLUTION BOOK. 

(3) For 12 da. it is \ of 12, or $.002. 

(4) Adding, we have $.202, which is the interest on $1 for the 

given time and rate. 

(5) On $380 the interest is 380 times $.202 = $76.76. 

PROBLEM 194. 

What principal will in 7 yr. 4 mo., at 8%, amount to $749.70? 

Solution. 

(1) The interest on $1 for the given time and rate is $.58|. 

(2) $1 + $.58f = $1.58$, amount. 

(3) $749.70 -^ 1.58* = $472.50, the required principal. 

PROBLEM 195. 

The sum of A's and. B's money on interest for 4 yr. 6 mo., at 6%, 
gives $5400 interest: how much money has each, if 3 times B's equals 
A's? 

Solution. 

(1) The interest on $1 for the given time and rate is $.27. 

(2) If $1 gives an interest of $.27, to give $5400 interest it 

will require as many dollars as .27 is contained times 

in $5400, or $20000. 

(3) | = A's and B's money = $20000. 

(4) \ of $20000 = $5000, B's money. 

(5) |, A's money = 3 times $5000 = $15000. 

.-. A's money is $15000, B's is $5000. 

PROBLEM 196. 
In what time will $1800, at 4£%, give $1247.40 interest? 

Solution. 

(1) The interest on $1800 at 4|% for 1 yr. is $81. 

(2) If in 1 yr. the principal gives $81 interest, to give $1247.40 

interest it will require as many times 1 yr. as $81 is 
contained times in $1247.42, which is 15| yr., or 15 yr. 
4 mo. 24 da. 

Rule. — Divide the given interest by the interest of the principal at 
the given rate for one year. If the amount is given y subtract the 
principal from the amount to fiiid the interest', and then proceed as 
before. 

PROBLEM 197. 
At what rate will $13.25, in 8 yr. 10 mo. 18 da., give $7,062* interest? 



INTEREST. 91 

Solution. 

(1) The interest of $13.25 for 8 yr. 10 mo. 18 da. at 1% is 

$1.17704. 

(2) If the principal in the given time at 1% gives $1.17704 in- 

terest, to give $7.0625 interest it will require as many 
times 1% as $1.17704 is contained times in $7.0625, 
which is 6%. 

PROBLEM 198. 

On a sum borrowed at 6% per annum and loaned at 8% per annum, 
I realized a gain of $31.20 in 3 mo. and 18 da: find amount loaned. 

Solution. 

(1) The interest on $1 for 3 mo. 18 da. at 6% is $.018. 

(2) The interest on $1 for 3 mo. 18 da. at 8% is $.024. 

(3) $.024 — $.018 = $.006, gain on $1. 

(4) If $1 gives a gain of $.006, to give a gain of $31.20 it 

will require as many dollars as $.006 is contained times 
in $31.20, or $5200. 

.-. $5200 — amount loaned. 

PROBLEM 199. 

The sum of § of A's plus \ of B's money being - on interest for 8 yr. 
at 6% gives $960 interest: what has each, if \ of B's is 3 times § of A's? 

Solution. 

(1) The interest on $1 for 8 yr. at 6% is $.48. 

(2) Then, to give $960 interest, it will require as many dollars 

as $.48 is contained times in $960, which is $2000, sum 
of | of A's + \ of B's money. 

(3) | + \ = |; $2000 Xf= $3000, B's. 

(4) \ of $3000 = $1500; $1500 H- 3 = $500. 

(5) £ = $500; \ — -J of $500 = $250. 

(6) f = 3 times $250 = $750, A's. 
.-. A's money is $650, B's $3000. 

PROBLEM 200. 
A's money added to f of B's, which is to A's as 2 to 3, being put on 
interest for 6 yr. at 4%, amounts to $744: what has each? 

Solution. 

(1) The interest on $1 for the given time and rate is $.24. 

(2) $1 + $24 = $1.24, amount of $1. 

(3) $744 -r $1 24 = $600, A's money + f of B's. 

(4) § A's money + | B's = f = $600. 

(5) I = $ of $600 == $120. 



92 FAIRCHILD'S SOLUTION BOOK. 

(6) f = 3 times $120 = $360, A's money. 

(7) | of $360 = $240, which is as 2 : 3. 

(8) Then, f = $240. 

(9) J = ^ of $240 = $120. 

(10) |=3 times $120 = $360, B's money. 

.'. A and B have each 



PROBLEM 201. 

f of the cost of A's mill, increased by i of the cost of his house for 
2 yr. at 5% amounts to $4950: what was the cost of each, if f of the cost 
of the mill was only f as much as f of the cost of the house? 

Solution. 

(1) The interest on $1 for the given time and rate is $10. 

(2) $1 + $.10 = $1.10, amount of $1. 

(3) $4950 -f- $1.10 = $4500, f of the cost of A's mill + f of 

the cost of his house. 

(4) f = f of f , or /g of house. 

( 5 ) i + A = If = i °f the cost of mil1 + i of the cost of 

the house. 

(6) || = $4500. 

(7) ¥ l = 3V of $4500, or $125. 

(8) ff = 35 times $125 = $4375, cost of house. 

(9) I of $4375 = $3500, | of the cost of the house. 

(10) f of $3500 = $1000, f of I of the cost of the house. 

(11) 1 = $1000. 

(12) I = i of $1000 =$3331. 

(13) I cost of mill = 5 times $333^ = $1666|, cost of the mill. 

.-. The house cost $4375, the mill $1666f. 

PROBLEM 202. 

. The interest on the sum of A's and B's money for 3 yr. and 9 mo. 
at 8%, is $3213; f of A's money is equal to f of B's: how much has each? 

Solution. 

(1) The interest on $1 for the given time and rate is 30 ct. 

(2) If $1 gives an interest of $.30, to give $3213 interest it 

will require as many dollars as $.30 is contained times in 
$3213, or $10710. 

(3) f of A's money is equal to J of B's. 

(4) 1 of A's = \ of f, or f of B's. 

(5) f A's = 3 times f = f of B's. 

(6) Then f = B's, and f A's. 



INTEREST. 93 

(7) | _|_ | — _y. t A's and B's money. 

(8) ig- = $10710. 

(9) i = T V of $10710 = $630. 

(10) | = 8 times $630 == $5010, B's money. 

(11) | = 9 times $630 = $5670, A's money. 

.*. A's money is $5670, B's $5010. 

PROBLEM 203. 

The amount of a certain principal, in a certain time, at 5%, is $833, 
and the amount for the same time at 12%, is $1047.20: required the 
principal and time. 

Solution. 

(1) 12% — 5% = 7%, the difference in rates of interest. 

(2) Now, the principal and time are the same; the difference 

in amounts must have resulted from the difference in 
the rates of interest. 

(3) The difference of the amounts equals the interest at 7%. 

$1047.20 — 1833 = $211.20, the difference in amounts, 
or 7% of the principal for the time. 
(1) 7% = $214.20. 

(5) 1% =jf of $214.20, or $30.60. 

(6) 5% of the principal for the time = 5 X $30.60 = 8153. 

(7) 12% of the principal for the time = 12 X 830.60, or 

$367.20. 

(8) Now, $833 — $153, or $1047.20 - $367.20 = $680, the 

principal. 

(9) 5% of $680 for 1 year = $34. 

(10) 12% of $680 for 1 year = $81.61; since the principal at 
5% yields $34 in 1 year, and at 12%, $81.60 in 1 year, 
the money must have been loaned for as many years as 
$34 is contained times in $153, or $81.60 is contained 
times in $367.20, which is ±\ yr., or 4 yr. and 6 mo. 

.'. $680 = the principal, and 4 yr. 6 mo. the time. 

PROBLEM 204. 

The amount of a certain principal for 4 yr., at a certain rate per 
cent, is $3551, and for L9 yr., $6929$: required the principal and rate. 

Solution. 

(1) 19 yr. — 4 yr. = 15 yr., the difference of time. 

(2) The principal and the rate are the same, the difference of 

amounts, $6929J — $3551, or $3378^, is the interest on 
the principal for 15 yr. 



94 FAIR CHILD'S SOL UTION B O OK. 

(3) $3378f -M5 = 1225.25, the interest on the principal for 

1 yr. 

(4) Then 4 X 1225.25 = $901, the interest on the principal 

for 4 yr. 

(5) 19 times X $225.25 = $4279f , the interest on the principal 

for 19 years. 

(6) Now, $3551 — $901, or $6929f — $4279$ = $2650, the 

principal. 

(7) $2650, the principal loaned at 1% will earn $26.50 in 1 yr. 

(8) The principal at the required rate earned $225.25 in 1 yr., 

then it must have drawn as many times 1% as $26.50 is 
contained times in $225.25, or 8J. 

.'. $2650 == the principal, 8%% the rate. 



PROBLEM 205. 

A owes B $900, and has only $350 cash on hand; he proposes to pay 
a part of this money on the debt, and to pay the interest in advance at 
10% : for what sum was the note drawn? 

First Solution. 

(1) Let 100% be the face of the note. 

(2) Then, he paid 20% interest, or he paid in all $900 + 20% 

of face of note. 

(3) He also paid $300 + 100% of the face, and $900 with 20% 

of the note = $350 and the whole note. 

(4) Subtracting 20% from each of the two equalities, we have 

80% of note + $350 = $900, or 80% = $550. 

(5) 1% = -i-o of $550 = $6 875. 

(6) 100% = 100 times $6 875 = $687.50, face of note. 

Second Solution. 

(1) If he had paid $350 on the principal, there would have 

remained $900 — $350 = $550 of the principal unpaid. 

(2) By the condition of the problem the $350 includes the 

interest on the unpaid balance of the principal for 2 yr. 
at 10%. Now, this unpaid balance must be $550 -|- the 
interest on the unpaid balance for 2 yr. at 10%. 

(3) Let 100 parts = the unpaid balance. 

(4) T W will equal the interest and T W will represent the $550. 

(5) A°o = $550. 

(6) tU = sV of $550, or $-^-. 

(7) -HHS-, the unpaid balance — 100 times $- 5 /o°- = $687.50. 

.-.$687.50 = face of new note. 



INTEREST. 95 

PROBLEM 206. 

By lending- a sum of money at 4% and another sum at 5%, the total 
interest is $10; if the rates are changed the interest is $9.80: find the 
principal lent at each rate. 

Solution. 

(1) 5% — 4% = 1%, difference in rates. 

(2) $10 — $9.80 = $,20, difference. 

(3) 1% = $.20. 

(4) $.20 H- 1% = $20, difference in amounts. 

(5) ($10 + $9.80)-^(5% + 4%) = $220, whole amount. 

(6) $220 — $20 = $200, now, this = both sums — $20. 

(7) .'. the 1st sum must be \ of $200, or $100, and the 2d sum 

$100 + $20, or $120. 

PROBLEM 207. 

The interest of \ of A's -f- f of B's money for a certain time at 2% 
was to this sum as 9 to 250, and the amount of this interest for 25 times 
as long, at 10 times as great a per cent was $180: what was their money, 
if A's money was to B's as 1 to 3? 

Solution. 

(1) Since the interest on the sum loaned was to that sum as 

9 : 250, the interest was ^fa of that principal. 

(2) The interest of any principal for one year at 2% = ^ of 

it; then -jf - -i- ^, or f yr. (1 yr. 9 mo. 18 da.) is the 
time. 

(3) 25 times this period is 45 yr., and 10 times the rate is 20%. 

(4) The amount of $1 for 45 yr. at 20% = $10; hence, $180 -f- 

$10, or $18, is the interest on the part loaned at 2% for* 
1 yr. 9 mo. 18 da., which by the 1st condition is ? |- of 
the principal. 

(5) Hence, $18 4- jfa — $500 = J A's + I B's principal, or 

part loaned. 

(6) | of B's money = 2 times A's, because they are in the ra- 

tio of 1 to 3. 

(7) Hence, $500 is 2£ times A's money. 

(8) *500 -r %k — $ 20 °. A ' s money, and $200 X 3 = $600, B's 

money. 

.'. A's money is $200, and B's $600. 

PROBLEM 208. 

If f of A's money equals £ of B's and % of B's equals § of C's, and 
the interest of all their money for 4 yr. 8 mo., at 6%, is $13190; how 
much has each? 



96 FAIRCHILD'S SOLUTION BOOK. 

Solution . 

(1) The interest on $1 for the given time and rate is $.28. 

(2) Hence the principal is $15190 -f- .28 = $54250. 

(3) f of A's money = -f of B's. 

(-4) \ of A's money = \ of |, or § of B's. 

(5) f A's money = 3 times J = f of B's. 

(6) f = B's money. 

(7) \ of C's money = f of B's money. 

(8) \ of C's money = \ of § = f of B's. 

(9) f C's money = 5 times f = y- of B's. 
(10) t + ! + ¥ — W» their money = $54250. 
(ID tV = drs of $54250 = $232lff. 

(12) % = 81 times $232iff = $18859.44, A's money. 

(13) f| = 72 times $232-iff == $16663.95, B's money. 

(14) ff = 80 times $232|f| = $18626.61, C's money. 



II. TRUE DISCOUNT. 

73. True Discount is the difference between the amount of 
the debt and the present worth. 

74. The Present Worth of a debt payable at a future time 
without interest, is such a sum as, being on interest for the 
time at a certain rate, will amount to the debt. 

Note. — The true discount is the interest on the present worth for 
the time between the payment of the debt and the time it becomes due. 

PROBLEM 209. 

What is the present worth of $1206, due 5 yr. 8 mo. hence without 
interest, money being- worth 



'c : 



Solution. 

(1) $1 loaned at that time and rate amounts to $1.34. 

(2) One could pay as many dollars for a note of $1206 as 

$1.34 is contained times in $1206, or $900. 

.*. $900 is the present worth. 

PROBLEM 210. 

The interest on a sum of money for a certain time is $300 and the 
true discount is $240: find the sum of money. 



TRUE DISCOUNT. 97 

Solution. 

(1) The difference between the interest and true discount of 

any sum is always equal to the interest on the true dis- 
count for the same time and rate. .'. $300 — $240 = 
160, the difference = the interest and $240 the principal. 

(2) $60, the interest, is -£fa, or J of the principal. 

(3) Then, \ of the principal = $300, the interest. 

(4) |, or the principal = 4 times $300, or $1200. 

.". $1200 = the principal. 

PROBLEM 211. 

The true discount of a $52 note, having- 8 months to run, is $4; and 
at the same rate the discount on a note of $85 is $5: find the time of the 
latter. 

Solution. 

(1) The sum on which discount is found is $52 — $4 = $48. 

(2) The interest on $48 for 8 mo. at 1% is $.32. 

(3) $4 -f- .32 = 12^%, ra te of discount. 

(4) On the other note $85 — $5 = $80, the proceeds. 

(5) 12|% of $80 = $10, discount for one year. 

(6) Hence, the time is 5 -7- 10 = 6 mo. 

.'. The time = 6 mo. 

PROBLEM 212. 

The true discount on a sum of money, at 6%, is $50 more than the 
sum of the true discounts on half the sum at 8%, and on the other half 
at 4% : find the sum. 

Solution. 

(1) The true discount on any sum is equal to the present 

worth of its interest for the same time and rate. 

(2) Hence, at 6% the true discount on the entire debt is yf 

of the debt; on half of the sum at 8% it is y-J-g-; on the 
other half at 4% it is yfj of it. 

(3) .'. [ T §8 - (t*s + tIt)] = tAW or this equals $50. 

(4) Then, the sum is as many dollars as frffiia is contained 

times in $50, or $148824. 

,\ $148824 is the required sum. 

PROBLEM 213. 

The sum of the simple interest, true present worth and true dis- 
count of a certain principal for a certain time is $5000: find the amount. 



98 FAIRCHIL&S SOLUTION BOOK. 

Solution. 

(1) The interest = Prt; the present worth = P -7- (1 + rt); 

and the discount = P — P + (1+rt) = Prt -7- (1+rt). 

(2) Then Prt + P 4- (1 + rt) + Prt -r- (1 + rt) = $5000. 

(3) Now, P -f- (1 + rt) + Prt -r- (1 '+ rt) = P(l + rt) -r- 

(1 + rt) - P. 

(4) /. We have Prt + P = $5000. 

(5) But P(l + rt) = A, amount, and A = $5000. 

.'. $5000 is the amount required. 

PROBLEM 214. 

The interest of a certain sum is 20% of it, and the true discount is 
$20: find the sum. 

Solution. 

(1) 100% = sum. 

(2) 100% -f- 1.20 = 83J%, present worth. 

(3) 100% — 83£% = 16f%,true discount. 

(4) .-. 16f % = $20. 

(5) 1% = $20 4- 16| = $1.20. 

(6) 100% = 100 times $1.20, or $120. 

.-. $120 = the sum. 

PROBLEM 215. 

The present worth is $100 more than the discount at 6%, and $95 
more at 7% : find the debt and time. 

Solution. 

(1) Let P = debt, t = time, r = 6%, and r' = 7%, 

(2) Then the present worth at 6% is P -7- (1 + rt) and at 7%, 

Pv (1 + r't). 

(3) P — [P -r- (1 + rt) ], or Prt -r (1 + rt) is the discount 

at 6%. 

(4) Then[P-^ (1 + rt)] — [Prt -r- (1 +rt)] = $100. 

(5) Whence t = (P — $100) -*- (P -f- $100) r .... (1). 

(6) P-[P-f (1 + r't)], or Pr't -^ (1 + r't) is the discount 

at 7%. 

(7) Hence, [P -r- (1 + r't)] — [Pr't -r- (1-f r't)] = $95. 

(8) t = (P — $95) -T- (P + $95)r' (2). 

(9) Substituting in these equations the values of r and r' and 

equating the values of t, we find P = $135.25 nearly, 
and from (1) t = 2.4969 yr. 

.'. The debt is $135.25, and the time 2.4969 yr. 



TRUE DISCOUNT. 99 

PROBLEM 216. 

What sum is it whose true discount by simple interest for 4 yr. is 
$25 more at 6% than at 4% per annum? 

Solution. 

(1) The true discount on any sum is equal to the present 

worth of its interest for the same time and rate. 

(2) Hence, the interest on $1 at 4% for 4 yr. = $.16. 

(3) On $1 at 6% for 4 yr. = $.24. Then $25 -r- Gr& — tVf = 

$449.50. 

.-. $449.50 = the sum. 

PROBLEM 217. 

Two-thirds of A's money is $100 less than \ of B's; the present 
worth of A's for 1 yr. at 1% is $200 more than the simple interest of 
B's for 2 yr. at 8%: find the present worth of B's for 1 yr. 7 mo. and 
19 da. at 9%. 

Solution. 

(1) If f of A's money + $100 = \ of B's, then f of A's + 

$200 = B's. 

(2) The true present worth of A's money for 1 year at 7% 

-J-§4 of A's money. 

(3) The interest on B's money for 2 yr. at 8% is -fe of (J of 

A's + $200) = l 7 f of A's money + $32. 

(4) Hence, fj4 of A's money — $200 = || of A's + $32. 

(5) m - \l of A's = $200 + $32. 

(6) A's is $232 -r- -fifl = $321.67. 

(7) B's is J of $321.67 + $200 = $628.89. 

(8) $1.14725 is the amount of $1 for 1 yr. 7 mo. 19 da. at 9%. 
<9) The present worth of $628.89 is $628.89 -f- $1.14725 = 

$548.17. 



is 



.'. $548.17 = present worth of B's money. 



PROBLEM 218. 

A note of $2000, dated July 4, 1876, due May 1, 1878, and bearing 
interest at 8%, was cancelled October 2."), 1877, by payment of the pres- 
ent worth at 6%: what was the present worth at this date, and the 
discount? {R. P., p. 259, prob. 75.) 

Solution. 

(1) The interest on $1 from July 4, 1876, to May 1, 1878, is 

$.146. 

(2) $2000 X $.146 = $292, interest on the note for the time. 

(3) $2000 + $292 = $2292, amount of note. 



100 FAIRCHILD'S SOLUTION BOOK. 

(4) The interest on $1 from Oct. 25, 1877, to May 1, 1878, is 

$.031. 

(5) $1 + $.031 = $1,031, the amount of $1 for time and rate. 

(6) Then there would be as many dollars in the present 

worth of the note as $1,031 is contained times in $2292, 
or $2223.08. 

(7) $2292 — $2223.08 = $68.92, the discount. 

.-. The present worth was $2223.08, the discount $68.92. 
PROBLEM 219. 

The true discount of a sum for 6 mo. is $5, and the interest for the 
same time and rate is $5.25: find the debt. 

Solution. 

(1) The difference between the interest and true discount of 

any sum is always equal to the interest on the true dis- 
count for the same time and rate. 

(2) $5.25 — $5 = $.25, the difference = the interest and $5 

the principal. 

(3) $.25, the interest, is 3^, or -^ of the principal. 

(4) f-g-, or the principal, is 20 times $5.25, or $105. 

(5) Suppose we were to find the rate, then we would find the 

interest on $105 for 6 mo. at 1%, or $.525. 

(6) $5.25, the interest on the principal = $5.25 -j- .525 = 10%. 

.'. The rate is 10% and the debt $105. 



III. BANK DISCOUNT. 

75. A Bank is an incorporated institution which receives 
and loans and money. 

76. Bank Discount is the interest on the face of the note 
for the time from the day of discount to the day of payment. 

77. The Proceeds of a note is the sum received for it when 
discounted, and equals the face less the discount. 

78. Term of Discount is the number of days from the time 
of discounting to the time of maturity of the note. 

Note 1. — The difference between bank discount and true discount 
may be shown as follows: If I take a note to the bank promising- to 



BANK DISCOUNT. 101 

pay $108 at the end of 1 yr., to get it cashed, by true discount I would 
receive $100; but by bank discount, not counting days of grace, I would 
receive $108 minus the interest of $108 for 1 yr., i. e., $108 — $8.64 = 
$99.36. 

Note 2. — The discount of an interest bearing note is computed on 
the amount of the note at its maturity. Banks compute interest for 
the actual number of days a note has to run. 

Note 3. — When no date of discount is given, the date of the note 
is taken as the date of discount. 

PROBLEM 220. 

A note of $200 is dated Aug. 25, payable in 60 days, and discounted 
at 6%: what is the proceeds? 

Solution. 

(1) We find the interest of $1 for 60 da. + 3 da., or 63 da., is 

$.0105, the bank discount. 

(2) $200 X .0105 == $2.10, bank discount. 

(3) $200 — $2.10 = $197.90, proceeds. 

PROBLEM 221. 

A note of $500 was given Jan. 1, 1895, at 8% interest, due in 6 mo.; 
it was discounted in bank March 1 at 10%: what was the proceeds of 
the note? 

Solution. 

(1) Counting forward, we have June 4, 1895, as the date of 

maturity; from March 1 to June 4 is 94 da., the time 
to run. 

(2) As this is an interest bearing note, the sum to be dis- 

counted is the amount of the note. 

(3) $.0406| = interest on $1 for 6 mo. 3 da. 

(4) $500 X .0406| = $20J, interest on the note. 

(5) $520-J- is the sum to be discounted; $.02fJ is the bank dis- 

count on $1 for the time. 

(6) $520J X .97 T V, the proceeds on $1 = $506.75, the proceeds 

of the note. 

PROBLEM 222. 
I had a 6% bond of $800, dated Jan. 1, 1896, due Jan. 1, 1897; on 
July 1, 1896, I sold the bond to Mr. Huntsman in such a way as to give 
him 8% on his investment; if Mr. Huntsman borrowed the money 
needed to pay the note, from a bank, at 10% for 90 da., what was the 
face of the bank note? 

Solution. 

(1) From Jan. 1, 1896, to Jan. 1, 1897, is 1 yr. 

(2) $48 = interest on $800 for 1 yr. at 6%, and $848 = am't 

due Jan. 1, 1897. 



102 FAIRCHILD'S SOLUTION BOOK. 

(3) Amount of $1 at 8% for 6 mo. — $1.04. 

(4) $848 -=- 1.04 = $815 T 5 3, paid for bond July 1, 1896. 

(5) 93 da. = time to run; the proceeds of $1 for 93 da. at 10 

% = $.974J, 

(6) .'. The face of the note must be as many dollars as .974| 

is contained times in $815 T 5 3, or $837,007. 

PROBLEM 223. 

A banker discounts a note at 8% per annum, thereby getting- 9% 
per annum interest: how long does the note run? 

Solution. 

(1) The discount is reckoned on the face of the note, but the 

interest is estimated on the proceeds. 

(2) Hence, 9%, or -fa of the proceeds == 8%, or -^ of the 

face. 

(3) y^ of the proceeds = -J- of -fe = -fa of the proceeds. 

(4) f{Hr, the proceeds = 100 times -fa = -|ff of the face. 

(5) The discount for the required time is |-|f — |~Jf = i of 

the face. 

(6) .'. The time is £ -^ ^ = ff yr., or 500 da. 

PROBLEM 224. 
A note dated February 19, 1876, payable January 1, 1877, and bear- 
ing 8% interest, was discounted October 12, 1876, at 6%; the proceeds 
was $1055.02: what was the face of the note? (R. jd R,p. 256.) 

Solution. 

(1) If the note is nominally due Jan. 1, 1877, it will be 

legally due 3 days later, or Jan. 4, 1877. 

(2) The interest on $1 from Feb. 19, 1876, to Jan. 4, 1877, is 

$.07; $1 + $.07 = $1.07, amount. 

(3) From Oct. 12, 1876, to Jan. 4, 1877, is 84 da., time to run. 

(4) $1.07 = amount of $1, or sum to be discounted; $.014 = 

bank discount on $1 for the time to run. 

(5) $1.07 X .014 = $.01498, bank discount on $1.07. 

(6) $1.07 — $.01498 = $1.05502, proceeds of $1 of the face of 

the note. 

(7) Then there would be as many dollars in the face of the 

note as $1.05502 is contained times in $1055.02, or 
$1000. 
.-. The face is $1000. 

PROBLEM 225. 
A commission merchant sold a consignment of cotton for $4500, 
receiving in payment a note, which yielded, on being discounted, 
$4475.25: what was the time of the note? {Brooks' Arith.) 



BANK DISCOUNT. 103 

Solution. 

(1) $4500 — $4475.25 = $24.75, the discount. 

(2) The discount on $1 for 1 day at 6% is $.000^, and on 

$4500 it is $4500 X .000^ = $.75. 

(3) Hence, the note is discounted for as many days as $.75 

is contained times in $24.75, or 33 days. 

(4) .'. The time is 33 da. — 3 da. or 30 da. 

Note. — When we are to find the actual time, the days of grace 
should not be subtracted. 

PROBLEM 226. 

Mr. Herr buys goods to the amount of $4000, and to pay for them 
gets his note for 60 days discounted at a bank; if the face is $4042.45, 
what is the note? {Brooks' Arith.) 

Solution. 

(1) $4042.45 — $4000 = $42.45, discount. 

(2) The discount on $4042.45 at 1% for 63 da. is $7,074. 

(3) Hence, the required rate is as many times 1% as $7,074 

is contained times in $42.45, which is 6%. 



SOME INTERESTING PROBLEMS. 

PROBLEM 227. 
A father left his four sons, whose ages were respectively 5, 9, 13, 
and 17 years, $27500, to be divided in such a manner that the respective 
shares being placed out at 5% simple interest, shall amount to equal 
sums when they become 21 years of age: what were the shares? 

{Putnam Co.) 
Solution. 

(1) It will be 16, 12, 8 and 4 years respectively until they 

are of age. 

(2) Now, the present wor;h of $1 for 16 years at 5% is $.55|. 

(3) The present worth of $1 for 12 yr. at 5% is $.62£. 

(4) The present worth of $1 for 8 yr. at 5% is $71f 

(5) The present worth of $1 for 4 yr. at 5% is $.83f 

(6) $.55f + $.62-i +$-71f+$.83j = $2.72}ff, the sum of 

the present worths. 

(7) $27500 -^ 2.72] 2 1 = $10080. 

(8) 1st son's share = $10080 X .55f = $5600. 

(9) 2d son's share = $10080 X .62-3, = $6300. 



104 FAIRCHILD'S SOLUTION BOOK. 

(10) 3d son's share = $10080 X .71f = $7200. 

(11) 4th son's share = $10080 X .83fc == $8400. 

PROBLEM 228. 
A note of $312 given April 1, 1872, 8% from date, was settled July 1, 
1874, the exact sum due being- $304.98. Indorsed: April 1, 1878, $30.96; 

October 1, 1873, $ ; April 1, 1874, $20.40. Restore the lost figures of 

second payment. 

Solution. 

(1) The new principal April 1, 1873, is ($312 X 1.08) — 

$30.96 = $306. 

(2) The new principal April 1, 1874, is $304.98 -f- $1.02 = 

$299. 

(3) We should ascertain whether the payment Oct. 1 met 

the interest due. If it did, then we find the new prin- 
cipal for Oct. 1 and subtract it from the amount due on 
that date; but if the payment was less than the inter- 
est, we can readily restore the lost payment by sub- 
tracting the new principal April 1, 1874, $299, from the 
amount then due, less the payment, $20.40. 

(4) The interest on $306 from April 1, 1873, to Oct. 1, at 8% 

is $306 X .04 = $12.24, and the amount due then would 
be $318.24. 

(5) The new principal Oct. 1 would be ($299 + $20.40) -7- 

1.04 = $307.11. 

(6) Hence, the lost payment could not be more than $318.24 

+ 1.04 = $307.11. 

(7) The lost payment could not be more than $318.24 — 

$307.11 = $11.13, which is less than the interest, $12.24, 
then due, and we must use the principal, $306, to April 
1, 1874. 

(8) The amount of $306 from April 1, 1873, to April 1, 1874, 

is $306 X $1.08 = $330.48. 

(9) Subtracting the payment, we have $330.48 — $20.40 = 

$310.08, the new principal, had nothing been paid in the 
meantime. 
(10) But we found the new principal to be $299, and the lost 
payment is $310.08 — $299 = $11.08. 

.'. The missing figures are $11.08. 
PROBLEM 229. 
Burt owed in two accounts $487; neither was to draw interest till 
after due, one standing a year and the other two years. He paid both 
in 1 yr. 5 mo., finding the true discount of the second, at 6%, exactly 
equal to the interest of the first: what difference of time would the 
common rule have made? 



SOME INTERESTING PROS IE MS. 105 

Solution. 

(1) One account drew interest 5 mo. and the other was dis- 

counted 7 mo. before due. 

(2) Interest on $1 for 5 mo. at 6% is 2% ct. = $^. 

(3) Interest on $1 for 7 mo. at 6% is 3^ ct. 

(4) Hence, the 2d account was bought for 96-^ ct. on the 

dollar. 

(5) The true discount on any sum is equal to the present 

worth of its interest for the same time and rate; hence, 
31 ct. -r 1.03-J = $*fo. 

(6) Now, the interest and discount are equal; .\ ^ = ^fe, 

and H = 40 X jfa, or fff 

0) m + m = m = $ 48 ?- 

(8) |-84 = $207, the 2d, and fff = $280, the 1st account. 

(9) $280 X 1 = $280, and $207* X 2 = $414. 

(10) ($280 + $414) -f- ($280 + 1207) = 1 yr. 5 mo. 3 da. 

(11) 1 yr. 5 mo. 3 da. — 1 yr. 5 mo. = 3 da. 

.*. The difference is 3 days. 

PROBLEM 230. 

A Cincinnati manufacturer receives, April 18, an account of sales 
from New Orleans; net proceeds $5284.67, due June 4-7. He advises his 
agent to discount the debt at 6%, and invest the proceeds in a 7 day 
bill on New York, interest off at 6%, at i% discount, and remit it to 
Cincinnati. The agent does this, April 26. The bill reaches Cincin- 
nati May 3, and is sold at \% premium. What is the proceeds, and how 
much greater than if a bill had been drawn May 3, on New Orleans, 
due June 7, sold at \% premium, and interest off at 6%? (R. H. A.) 

Solution. 

(1) From April 26 to June 7 is 42 da. 

(2) The discount on $1 for 42 da. at 6% is $.007. 

(3) $1 —$.007 = $.993, proceeds. 

(4) $5284.67 X .993 = $5247.6773, amount invested in the 

bill on N. Y. 

(5) 7 da. + 3 da. of grace = 10 da. 

(6) \% = discount; 100% — -\% = 99|%. 

(7) Bank discount of $1 for 10 da. is -J-%. 

(8) 99^% — i% = 99£%, or $.99^ 

(9) The N. Y. bill = $5247.6773 -r- .99J = $5282.896. 

(10) In Cincinnati the bill sold at $1.00^ on the dollar, or 

$5282.896 X 1.00J = $5296.10, proceeds. 

(11) From May 3 to June 7 is 35 da. 



106 FAIRCHIL&S SOLUTION BOOK. 

(12) \°] — premium; the bank discount on $1 for 35 da. at 

6% is $.005£. 

(13) Then, the New Orleans bill is worth $5284.67 X ($1 — 

$.006f + $.001) — $5260.45. 

(14) .-. The gain = $5296.10 — $5260.45 = $35.65. 

PROBLEM 231. 

What rate of income do I realize by purchasing- United States 4% 
bonds at 105, if I sell them in six years at 104? 

Solution. 

(1) .04 X 6 = .24. 

(2) 1.04 + .24 = 1.28, amount realized on bond. 

(3) 1.28 — 1.05 = .23, amount gained in 6 yr. 

(4) .23 ■—- 6 — .03f, amount gained in 1 yr. 

(5) .03f -i- 1.05 = m%, rate of gain. 

Note.— Solved by Prof. G. B. M. Zerrforthe Mathematical Monthly. 
PROBLEM 232. 

A man boug-ht a farm for $5000, agreeing- to pay principal and in- 
terest in five equal annual installments: what is the annual payment, 
including- interest at 6%? {Milne.) 

Solution. 

(1) The interest on $5000 for 1 yr. at 6% is $300. 

(2) ($300 X 1.3382256) ~ .3382256 = $1186.08, one of the 

equal payments. 

Rule. — Multiply the interest on the debt for 1 year by the com- 
pound amount of $1 for the given time and rate, and divide the 
product by the compound interest for the same time and rate. 

PROBLEM 233. 

A miller buys a mill for $6000, agreeing- to pay for it in three equal 
annual payments, he paying- 6% on the debt: .what payment does he 
make? 

Solution. 

(1) The interest on $6000 for 1 yr. at 6% is $360. 

(2) Hence, $360 -^ (1.191016 — 1) = $1884.65, the part of 

the principal paid in 1st installment. 

(3) Now, the compound amount of this for 3 yr. at 6% is 

$1884.65 X 1.191016 = $2244.65, the payment required. 

Rule. — To find the part of principal paid in first installment, 
divide the interest on the debt for 1 year by the compound interest for 
the given time. 



CHAPTER XI. 

COUNTY EXAMINATION TESTS, AND OTHER 
PROBLEMS. 

PROBLEM 234. 

An agent sold a lot of cotton on commission of 4% ; he invested the 
net proceeds in grain, after keeping his commission of 3% : if If times 
his commission is $3 more than $340, what was the value of the grain? 

{Putnam Co.) 
Solution. 

(1) f = his commission. 

(2) {Xl} = { = J3 more than $340, or $343. 

(3) i = l r of $343 = $49. 

(4) f , his commission = 5 times $49, or $245. 

(5) Out of every $1 receipts of cotton the agent receives 

4 ct. commission; 96 ct. = net proceeds. 

(6) T J ¥ of 96 ct. = ? § § ct., 2d commission. 

(7) 4 ct. + ■}-§-$ ct. = f-jJ-J ct., or § T i?> whole commission. 

(8) $t5~j = $245. Then, there must have been as many dol- 

lars of receipts for cotton as $ T ^ is contained times in 
$245, or $3605. 

(9) Now, as $ 1 j 8 is the whole commission on $1, r J ¥ of 

$3605 = $245, whole commission. 
(10) $3605 — $245 = $3360. 

.'. $3360 is the value of the grain. 

PROBLEM 235. 

A, B and C own a factory, a mill, a foundry respectively; the fac- 
tory is worth 8% less than the mill, and the mill 32% more than the 
foundry; C has traded the foundry for 75% of the factory, thus losing 
$178.40: what is the value of the mill? {Putnam Co.) 

Solution. 

(1) 100% = value of the mill (A). 

(2) 100% — 8% = 92%, value of the factory. 

(3) 100% = value of the foundry. 
(107) 



108 FAIRCHILD'S SOLUTION BOOK. 

(4) 132% = value of the mill in terms of the foundry. 

(5) 132% = 100%, the value of the mill from ... (A). 

(6) 1% = ^ of 100% = f| %. 

(7) 100% = 100 X |f = 75§f %, value of the foundry. 

(8) 75% of 92% = 69%, what C received for the foundry. 

(9) 75ff % — 69% = 6f|%, what C lost in the trade. 

(10) $178.40 = what he lost. 

(11) 6-JHj- % = $178.40. 

(12) 1% = $178.40 -=r 6ff = $26.40. 

(13) 100% = 100 times $26.40 = $2640. 

.-. The mill is worth $2640. 

PROBLEM 236. 
A sold a sheep and lost 25% ; if he had paid $1 more for it, he would 
have lost 40% : what did he pay for the sheep? {Montgomery Co.) 

Solution. 

(1) 100% = actual cost of the sheep. 

(2) 25% of 100% =25%, loss. 

(3) 100% — 25% = 75%, selling price of the sheep. 

(4) 100% + $1 = supposed cost of the sheep. 

(5) 40% of (100% + $1) = 40% + $.40, loss. 

(6) (100% + $1) — (40% + $.40) = 60% + $.60, selling 

price. 

(7) 75% = 60% + $.60. 

(8) 15% = $.60. 

(9) 1% = T v f $.60 = $.04. 

(10) 100% = 100 times $.04 = $4, the actual cost. 
.*. He paid $4 for the sheep. 

PROBLEM 237. 
A, B and C can do a piece of work in 4 days, A and C in 8 days, B 
and C in 6 days: how long will it take each working- alone? ( Wood Co.) 

Solution. 

(1) 4 da. = time it takes A, B and C to do the work; \ = 

part they do in 1 da. 

(2) 8 da. = time it takes A and C to do the work; -§- — part 

they do in 1 da. 

(3) 6 da. = time it takes B and C to do the work; £ = part 

they do in 1 da. 

(4) \ t part A, B and C do in 1 da. — -J, part A and C do in 

1 da. = -J-, part B does in 1 da. 

(5) | = part B does in f -f- -J- = 8 da. 



COUNTY EXAM IN A TION LISTS. 109 

(6) \, part A, B and C do in 1 da. — £, part B and C do in 1 

da. = 3^2, part A does in 1 da. 

( 7 ) tt = P art A does in ii i- A = 12 da y s - 

(8) -|-, part B and C do in 1 da. — \, part B does in 1 da. = Jj-, 

part C does in 1 da. 

(9) || = part C does in || -^- ^ = 24 da. 

.'. A can do the work in 12 da., B in 8 da., and C in 24 da. 



PROBLEM 238. 

I bought an article and sold it so as to g-ain 10% ; if it had cost 20% 
less, and I had sold it for one dollar less, I would have gained 25%: 
find the cost of the article. {Noble Co.) 

Solution. 

(1) 100% = actual cost. 

(2) 110% = actual selling price. 
(8) 80% = supposed cost. 

(4) 25% of 80% = 20%, gain. 

(5) 80%? cost + 20 % gain = 100%, supposed selling price. 

(6) But by the condition of the problem 110% — $1 = 100%. 

(7) 10% = |1. 

(8) 1% = T Vof$l = $.10. 

(9) 100% = 100 times $.10 = $10, the actual cost. 

.'. $10 is the cost of the article. 

PROBLEM 239. 

A banker had $1800, part of which he loaned at 6%, and the re- 
mainder at 5%, thus realizing- an income of $100: find the amount loaned 
at 6%. {Muskingum Co.) 

Solution. 

(1) 100% = amount loaned at 6%. 

(2) 6% of 100% = 6%, income on amount loaned at 6%. 

(3) $1800 — 100% = amount loaned at 5%. 

(4) 5% of ($1800 — 100%) = $90 — 5%, income on am't 

loaned at 5%. 

(5) 6% — ($90 — 2%) = difference of incomes. 

(6) 6% — ($90 — 2%) = $100. 

(7) 1% = $10. 

(8) 100% = 100 times $10 = $1000. 

.*. $1000 = amount loaned at 6%. 



110 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 240. 

Sold wheat at a loss of 5% and invested $25 more than the proceeds 
received from the wheat in corn, which I sold at a gain of 6% , making- 
a net g-ain of $8.50: find the value of the wheat. [Muskingum Co.) 

Solution. 

(1) 100% = value of the wheat. 

(2) 5% of 100% = 5%, loss. 

(3) 100% — 5% = 95%, proceeds. 

{4) 95% + $55 = amount invested in corn. 

(5) 6% of (95% + $25) = 5.70% + $1.50, gain. 

(6) 5.70% + $1.50 gain — 5% loss = $8.50, net gain. 

(7) .70% = $7.00 

(8) 1% = . T V of $7.00 = $10. 

(9) 100% = 100 times $10 = $1000. 

.-. $1000 = value of the wheat. 

PROBLEM 241. 

Paid $900 for a note due in 3 months, and sold it to a broker on the 
same day, taking bank discount at 6% , and cleared $84.50: find face of 
the note. , [Morrow Co.) 

Solution. 

(1) It is evident that $900 + $84.50, or $984.50, is the net pro- 

ceeds on the note. 

(2) The net proceeds on $1 for 3 mo. 3 da. at 6% is $.9845. 

(3) Then there would be as many dollars in the face of the 

note as $.9845 is contained times in $984.50, or $1000. 

.-. $1000 is the face of the note. 

PROBLEM 242. 

If A had $60 more money, he could buy 30 oxen, or with $120 less he 
could buy only 15: how much money has he? [York Co., Pa.) 

Solution. 

(1) 100% = his money. 

(2) (100% + $60) -T- 30 = cost of 1 ox. 

(3) (100% — $120) -f-15 = cost of 1 ox by second condition. 

(4) [(100% + $60) -7- 15] — [(100% — $120) -r 15] = 15 

oxen = $180. 

(5) $180 -r- 15 = $12, or the cost of 1 ox. 

(6) (100% + $60) -r 30 = $12, the cost of 1 ox. 

(7) 100% = $300, A's money. 

.-. A has $300. 



CO UNTY EX A MI N A TION LIS TS. Ill 

PROBLEM 243. 

On goods bought for $4500, on 6 months credit, I was offered 5% off 
for cash; if money was worth 6%, how much did I lose by accepting 
the credit? {Darke Co.) 

Solution. 

(1) 5% of $4500 = $225, amount deducted for cash. 

(2) $4500 — $225 = $4275, amount required for cash. 

(3) The interest on $1 for 6 mo. at 6% is $.03, and $4275 is 

$4275 X $.03, or $128.25. 

(4) $4275 + $128.25 = $4403.25. 

(5) $4500 — $4403.25 = $96.75, amount lost accepting the 

credit. 

.-. I lost $96.75. 

PROBLEM 244. 

Kerosene is bought at 50 ct. a gal., 10% of it is wasted: at what 
price must it be offered in order that the price may be discounted 10%, 
and yet 10% be made on the investment? {Hancock Co.) 

Solution. 

(1) Let 100 represent the number of gallons. 

(2) 100 X 50 ct. = $50, the cost of the kerosene. 

(3) 10% = gain; 10% of $50 = $5, gain. 

(4) $50 + $5 = $55, selling price. 

(5) 100% = marked price; 10% = deduction. 

(6) 100% — 10% = 90%, selling price = $55. 

(7) 1% = £> of $55 = $| |f. 

(8) 100% = 100 X $y = $6l£, marked price. 

(9) 10% of 100 gallons = 10 gallons, amount wasted. 

(10) 100 — 10 = 90 gallons, amount remaining. 

(11) $61 J- -r 90 == 67|-f ct., marked price per gallon. 

.'. It must be offered at 67$f ct. per gallon. 

PROBLEM 245. 

Sold a horse and a cow for $230; on the horse I gained 25%, and 
lost 25% on the cow: what was my gain, if the cow cost | as much as 
the horse? {Hardin Co.) 

Solution. 

(1) 100% = cost of the horse; 25% = gain. 

(2) 100% + 25% = 125%, selling price of the horse. 

(3) | of 100% = 66}%, cost of the cow. 

(4) 25% of 66}% = 16}%, loss. 

(5) 66}% — 16}% = 50%, selling price of the cow. 



112 FAIR CHILD'S SOLUTION BOOK. 

(6) 125% + 50% = 175%, selling price of both. 

(7) $210 = selling price of both. 

(8) 175% = $210. 

(9) 1% = T 1 tT of $210 = $1.20 

(10) 100% = 100 times $1.20 = $120, cost of the horse. 

(11) 66f % = 66f times $1.20 == $80, cost of the cow. 

(12) $120 + $80 = $200, cost of both. 

(13) $210 — $200 = $10, gain. 

.-. My gain is $10. 

PROBLEM 246. 

A broker charged me 2^% for purchasing- some uncurrent bank 
notes at 15% discount; three bills of $20, $50 and $100 respectively, 
turned out to be worthless, but by selling- the rest at par, I made $85: 
what was the face of the notes? {Logan Co.) 

Solution. 

(1) 100% = face; 100% — 2±% = 97J%, the proceeds. 

(2) 85% = the cost; 97|% —85% = 12f%, gain. 

(3) $20 + $50 + $100 = $170, loss. 

(4) But we see that he made $85 clear, then to make up the 

loss of $170 he must gain $170 + $85, or $255. 

(5) 12f % = $255. 

(6) 1% = $255 -T- 12| = $20. 

(7) 100% = 100 times $20 = $2000. 

.;. $2000 = the face. 

PROBLEM 247. 

At a certain time between 8 and 9 o'clock a boy stepped into the 
schoolroom, and noticed the minute hand between 9 and 10; he left, and 
on returning within an hour, he found the hour hand and minute hand 
had exchanged places: what time was it when he first entered the 
room, and how long was he gone? {Hancock Co.) 

Solution. 

(1) | = distance minute hand moves, or the distance it was 

beyond the hour hand. 

(2) %£- = distance the minute hand moved while the hour 

hand traveled f . 

(3) \H| = H~y distance they both moved. 

(4) .'. \ 6 - = 60 min. 

(5) \ = & of 60 = If, or 2 T \ min. 

(6) | = 2 times 2^ = 4 T 8 g min., the distance the minute 

hand was in advance of the hour hand. 



COUNTY EXAMINATION LISTS. 113 

(7) -§ = distance the hour hand moved past 8. 

(8) ^ = distance the minute hand moved. 

(9) \ 4 - = 40 min - + *A min - + i. or 44 A min + f • 

(10) -\ 2 - = 44^ min. 

(11) \ = A of 44A- = 2^ min. 

(12) -^ — 24 times 2^ min. = 48 T \% min., past 8. 

(13) Now, if the hands changed places, the minute hand fell 

short 4y^ min. of being an hour. 

.-. It was 48-j^- minutes past 8 when he stepped into the 
room, and he was gone 55 T \ minutes. 

PROBLEM 248. 

A and B wish to sell their horses to C; A asks a certain price, and 
B 50% more than A; A then reduces his price 20%, and B reduces his 
price 30%; C takes them* both, paying- $148: find the asking price of 
each. {Scioto Co.) 

Solution. 

(1) 100% = A's asking price. 

(2) 150% = B's asking price. 

(3) 20% of 100% = 20%, amount deducted from the asking 

price of the 1st horse. 

(4) 100%, — 20% = 80%, selling price of the 1st horse. 

(5) 30% of 150% = 45%?, amount deducted from the asking 

price. 

(6) 150% — 45%? = 105%, selling price of^2d horse. 

(7) 105% + 80% = 185%, selling price of both horses. 

(8) 185% = $148. 

(9) 1% = T -i-^ of $145 = $.80. 

(10) 100% = 100 times $.80 = $80, asking price of 1st horse. 

(11) 150% = 150 times $.80 = $120, asking price of 2d horse. 

/. A first asked $80, B $120. 

PROBLEM 249. 

A and B are in partnership, A investing $400 and B $500: how 
much must A put in at the end of two months to entitle him to half of 
the year's profits? 

Solution. 

(1) A has $400 X 12 = $4800 for 1 month. 

(2) B has $500 x 12 = $8000 for 1 month. 

(3) .-. A must contribute $6000 — $4800 = $1200 for 1 month. 

(4) For 10 months he must contribute $1200 -r 10, or $120. 



114 FAIRCHIL&S SOLUTION BOOK. 

PROBLEM 250. 

A sold goods which cost him $300, at a certain rate of profit; B sold 
the same goods to C at the same rate of profit; C paid $432 for them: 
what did B pay for them? {Ross Co,) 

Solution. 

(1) 100% = cost of goods to B. 

(2) r = rate of profit to A and B. 

(3) 100% + r = selling price of B's goods. 

(4) (100% + r) times $300 = $300 + 300r, selling price, or 

B's cost. 

(5) (100% + r) ($300 + 300r) = $300 + 600r + 300r*, B's 

selling price, or C's cost. 

(6) $432 = C's cost. 

(7) .-. $300 + 600r + 300r* = $432. 

(8) r* + 2r = «f. 

(9) r = .20, or 20%, A's and B's rate of gain. 

(10) Then, $300 X 20% = $60, A's gain. 

(11) $300 + $60 = $360, A's selling price. 

.'. B paid $360. 

PROBLEM 251. 

A farmer sold two horses for $810, receiving | as much for the first 
as for the second; on the first he gained 33&%, on the second he lost 
11£% : how much did he gain? (Brown Co.) 

Solution. i 

(1) 100% = selling price of 2d horse. 

(2) | of 100% = 80%, selling price of 1st horse. 

(3) 100% + 80% = 180%, selling price of both. 

(4) 180% = $810, selling price of both. 

(5) 1% = y^ of $810 = $4.50. 

(6) 100% = 100 times $4.50 = $450, selling price of 2d horse. 

(7) 80% = 80 times $4.50 = $360, selling price of 1st horse. 

(8) 100% = cost of the 1st horse. 

(9) 100% + 33J% = 133%, selling price of 1st horse. 

(10) 133i% = $360. 

(11) 1% = $360 -f- 133J- = $2.70. 

(12) 100% = 100 times $2.70 = $270, cost of 1st horse. 

(13) 100% — 1H% = 88f %, selling price of 2d horse. 

(14) .-. 88f% = $450. 

(15) 1% = $450 -^ 88f = $5.0625. 



COUNTY EX A MINA TION LISTS. 115 

(16) 100% = 100 times 15.0625 = $506.25, cost of 2d horse. 

(17) $270 + $506.25 = $776.25, cost of both. 

(18) $810 — $776.25 = $33.75, gain. 

.-. He gained $33.75. 

PROBLEM 252. 

Bought a certain number of egg's at 2 for a cent, and as many more 
at 3 for a cent; sold them all at 5 for 2 cents, and lost four cents: how- 
many eggs were there? 

Solution. 

(1) \ ct. = cost of 1 egg at 2 for a cent, 

(2) \ ct. = cost of 1 egg at 3 for a cent. 

(3) (-J ct. + \ ct.) ~ 2 = -f% ct., average cost of 1 egg. 

(4) | ct. = selling price of 1 egg. 

(5) A ct. — | ct. = - 6 J ct., average loss on each egg. 

(6) Hence, to lose 4 ct., there were as many eggs sold as ^ 

ct. is contained times in 4 ct., or 240. 

.*. 20 dozen eggs were sold. 

PROBLEM 253. 

A note is drawn May 20 on 6 months time for $840, bearing 6% int- 
erest; it is discounted at bank Sept 9, at 8% : find time to run and pro- 
ceeds. (Darke Co.) 

Solution. 

(1) Counting forward, we have Nov. 23 the date of maturity. 

(2) From Sept. 9 to Nov. 23 is 75 da., the time to run. 

(3) As this is an interest bearing note, the sum to be dis- 

counted is the amount of the note. 

(4) $.0305 = interest on $1 for 6 mo. and 3 da. 

(5) $840 X .0305 = $25.62, interest on the note, and $865.62 

= sum to be discounted. 

(6) $.01| = bank discount for the time and rate. 

(7) $^65.62 X .01f = $14,427, bank discount. 

(8) $865.62 — $14,427 = $851,193, proceeds. 

.'. The time to run is 75 da., and the proceeds $851,193. 

PROBLEM 254. 

A banker had $1800, part of which he loaned at 6%, and the re- 
mainder at 5%, thus realizing an income of $100: find the amount 
loaned at 6%. {Darke Co.) 



116 PA IP CHILD'S SOLUTION BOOK. 

Solution. 

(1) 100% = amount loaned at 6%. 

(2) $1800 — 100% = amount loaned at 5%. 

(3) 6% of 100% = 6%, interest on 100%. 

(4) 5% of ($1800 — 100%) = $90 — 5%, interest. 

(5) 6% + $90 — 5% = the whole income = $100. 

(6) 1% = $10. 

(7) 100% = 100 times $10 = $1000, amount loaned at 6%. 

(8) $1800 — $1000 = $800, amount loaned at 5%. 

/.He loaned $1000 at 6%. 

PROBLEM 255. 

A grain dealer sold a quantity of rye and wheat for $1320, gaining 
33^% on the rye and 12£% on the wheat: what was his total gain, if he 
received 20% more for the wheat than for the rye? (Putnam Co,) 

Solution. 

(1) 100% = selling price of the rye. 

(2) 120% = selling price of the wheat. 

(3) 100% + 120% = 220%, whole selling price. 

(4) 220% = $1320. 

(5) 1% = ^ of $1320 = $6. 

(6) 100% = 100 times $6 = $600, selling price of the rye. 

(7) 120% = 120 times $6 = $720, selling price of the wheat. 

(8) 100% = cost of the rye. 

(9) 133J% = selling price of the rye = $600. 

(10) 1% = $600 -=- 133! = $4 .50. 

(11) 100% = 100 times $4.50 = $450, cost of the rye. 

(12) 100% = cost of the wheat. 

(13) 112!% = selling price of the wheat = $720. 

(14) 1% = $720 -r 112! = *6.4&flfc. 

(15) 100% = 100 times $6.42 2 4 9 V» = $642^, cost of the wheat. 

(16) $600 — $450 = $150, gain on the rye. 

(17) $720 — $642-2^9 = $77||4, gain on the wheat. 

(18) $150 + $77|f! = $227|U . 

.'•. $227|fJ was his whole gain. 

PROBLEM 256. 

At what price per barrel shall an agent be ordered to buy potatoes 
at 2% commission that, after paying 7 ct. per baekel for transportation, 
they can be sold at $1.76 per barrel and net 10% profit? [Putnam Co.) 









COUNTY EXAMINA TION LISTS. 117 

Solution. 

(1) 100% = cost. 

(2) 110% = selling price. 

(3) $1.76 = selling price. 

(4) 1% =.- Tfra- of $1.76 = $.0160. 

(5) 100% = 100 times $.0160 = $1.60, cost to agent. 

(6) $1.60 — $.07 = $1.53. 

(7) 100 = number of parts investment. 

(8) 2 = number of parts commission. 

(9) 102 = whole number of parts. 
(10) U} of $ 153 = & 1 - 50 - 

.*. $1.50 = what the agent must pay per barrel. 

PROBLEM 257. 

A commission merchant received $4456.40 to invest in wheat, after 
deducting - his commission of 2%, and dray age 50 cts. per load of 36 bu.: 
how much wheat did he purchase at $1.20 per bushel? {Putnam Co.) 

Solution. 

(1) 100 = number of parts invested in wheat, and 2 — num- 

ber of parts commission. 

(2) 100 + 2 = 102, whole number of parts. 

(3) $4456.40 = whole number of parts. 

(4) \%% of $4456.40 = $4369.01, amount invested in wheat. 

(5) $.50 -=- 36 = $ Ol^g, cost of drayage per bushel. 

(6) .'. $1.20 -f $.01 T V = $121 T V cost per bushel. 

(7) $4369.01 -r- 1.2L& = 3599+ bushels. 

PROBLEM 258. 

A manufacturer gained 35% on one kind of wares and 44% on 
another: if the cost of manufacturing each kind was the same, and the 
gain was $360 more on the higher priced wares, what were the total 
rates? (Putnam Co.) 

Solution. 

(1) 100% = cost of each kind of wares. 

(2) 100% + 35% = 135%, selling price of the 1st kind. 

(3) 100% + 44% = 144%, selling price of the 2d kind. 

(4) 144% — 135% = 9%, gain on the 2d. 

(5) $360 = gain on the 2d. 

(6) 9% = $360. 

(7) 1% = I of $360 = $40. 

(8) 100% = 100 times $40 = $4000, cost of each kind. 



118 FAIRCHILD'S SOLUTION BOOK. 

(9) 135% = 135 times $40 = $5400, selling price of 1st kind. 
(10) 144% = 144 times $40 = $5760, selling price of 2d kind. 

.". The selling prices were $5400 and $5760 respectively. 

PROBLEM 259. 

A lady spent in one store \ of all her money and $1 more; in another, 
\ of the remainder and $1 more; in another, \ of the remainder and $1 
more; in another, \ of the remainder and $1 more; and in another, \ of 
the remainder and $1 more; she then had nothing- left: what sum had 
she at first? {Hancock Co. ) 

First Solution. 

(1) Let x = her money at first. 

x 

(2) -Q- + 1 = amount spent in 1st store. 

(3) x — ( -~- + 1 ) = -9 lj amount remaining. 

x x 

(4) -7 \ -\- 1, or -j + \ — amount spent in 2d store. 

X i X \ X 

(5) -g 1 —(-4- + i)= -y — i» amount remaining. 

x x 

(6) —~ — 1 + 1, or -Q- + \ — amount spent in 3d store. 

o o 

X ( X \ x 

(7) -j- — I — \-^ + i) = -g- — |, amount remaining. 

x x 

(8) r-^ — \ -{- 1, or r-^ + i = amount spent in 4th store. 

(9) -|- — i— ^ + i) = Ye~ V. amount remaining. 

(10) Now, by the condition of the problem, -^ = J^ 5 -. 

(11) x = $30, amount she had at first. 

Second Solution. 

(1) Starting with the last amount, or amount she must have 

had on entering the 4th store, which must have been 
$1 + $1, or $2; then, $2 + $1 = $3; $3x2 = $6, amt. 
on entering 3d store. 

(2) $6 + $1 = $7; $7X2 = $14, amount on entering 2d store. 

(3) $14 + $1 = $15; $15 X 2 = $30, amount on entering 1st 

store. 



CO UN TV EXAM IN A TION LISTS. 119 

PROBLEM 260. 

If the sun crossed the equator March 20, at 8:25 A. M., central stan- 
dard time, in what longitude did it cross? 

Solution. 

(1) Difference between standard and local time is 28 min. 

(2) 8 hr. 25 min. + 28 min = 8 hr. 53 min. 

(3) 8 hr. 53 min. = 133° 15'. 

(4) We live in about 84° west longitude. :. 133° 15' — 84° = 

49° 15' east longitude. 

Note. — When you pass from one section into the next, the time 
becomes one hour slower if you are moving - westward, one faster if you 
are moving eastward. Eastern time is that of the 75th meridian, Cen- 
tral time that of the 90th, Mountain time that of the 105th, Pacific 
time that of the 120th. See Art. 16. When it is noon at all places in 
the Eastern section, it is 11 a. m. at all places in the Central section, 
10 a. m. at all places in the mountain section, and 9 a. m. at all places 
in the Pacific section. This neat and simple system is now in use all 
over the United States and Canada. 

PROBLEM 261. 

If the earth's rotation were reversed, how many days would we 
have in a year? {Putnam Co.) 

So ht Hon . 

(1) If the earth's rotation were reversed we would have 367 

days in a year. The earth would make about 4 minutes 
less than one complete rotation each day; this would 
make a gain of one day in the year. 

(2) The earth makes 366 revolutions on its axis each year, 

and this would give us 367 days. 

PROBLEM 262. 

If the gain is VIV/r of the selling price: what is the rate of gain? 

{Putnam Co.) 
Solution. 

(1) Let 100% = selling price. 

(2) 12-^% of 100% = 12|%, gain. 

(3) 100% — 12J% = 87-|%, cost. 

(4) Let 100% = cost. 

(5) 87i% = 100%. 

(6) 1% = 100% -h 87| = i-l%%. 

(7) 12£% = 12| times 8 S°i%> or 14* %. 

.*. 142% is the rate of gain. 



120 



FAIR CHILD'S SOLUTION BOOK. 



PROBLEM 263. 

A owns the S. W. £ of the N. E. £, the S. E. i of the N. W. $, the 
N. E. i of the S. W. £, and the N. W. £ of the S. E. i of a section of 
land. Show by a drawing- the position of his farm in the section, and 
divide the remainder of the section into four farms of equal shape and 
area, giving- their dimensions. 

Solution. 

Let ABCD be the section 
of land, AB or AD = 320 rd. 

Divide the section into 
quarters containing 160 A. 
each. Then TOXD contains 
160 A., and ROYN 'contains 
40 A. 

It is obvious that A owns 
PSZN, or 160 A. 

We have left the four rect- 
angles WK, KL, LE and 
EW, each containing \ of 
(640 A. — 160 A.) = 120 A. 

DC = 320 rd.; XC = \ of 
320 rd., or 160 rd., and LC = 
| of 160 rd., or 80 rd. LC = 
KB; then CK = 320 rd. — 

80 rd. = 240 rd. .*. The rectangles are 240 rd. long and 
rd. wide. 




FIG. i. 



FIG. 2. 



PROBLEM 264. 

What will it cost to fence the N. \ of the E. i of the]N. E. \ of sec- 
tion 16, making- 5 rectangular fields of equal area, at $2 per rod? 

Solution. 

Let ABCD be the N. E. \ of sec- 
tion 16, FKBC = the E. \ of the N. 
E. \ of section 16, and OSCF the N. 
\ of the E. J of the N. E. \. 

OS, or OF = 80 rd., and the en- 
closed farm is 80 X 4 = 320 rd. 

Then divide it into two parts by a 

fence 80 rd. long, running east and 

west, and 32 rd. south; of north 

boundary; next divide the northern 

portion into two equal parts by a 

fence 32 rd. long, running north and 

south. Divide NOST into three equal parts by two fences 

(LX and EP) each 48 rd. long. This requires 320 + 80 + 32 

+ 48 X 2 = 528 rd. of fence. 




COUNTY EXAMINA TION LISTS. 121 

.-. 528 rd. X $2 = $1056, the cost of fencing. 

PROBLEM 265. 

My agent sold produce at 2% commission; increasing the proceeds 
by $42, I ordered him to purchase dry goods at 5% commission, which 
I sold at 3£% loss, my whole loss, including commission, being $63: 
find value of produce and dry goods. {Putnam Co.) 

Solution. 

(1) Out of every $1 received for produce, my agent received 

2 ct. commission, and secondly yf^ of 98 ct., or j-%% ct. 

(2) I lost 3| ct., or ffo of H* of 98 ct. = f-ff ct. 

(3) In all, my loss on $1 of produce = 2 ct. + f§£ ct. + 

«f ct.,or$W. 

(4) My loss on $42 is first ^ of $42 = $f^, and secondly 

m of m of $42 = $H£. 

(5) $H* + *H* = 9Hb total loss on $42. 

(6) Then, $63 — $B£ = $- 6 x W> !oss on the produce. 

(7) Now the loss on $1 of the produce is $ l fb 4 B a ; then there 

are as many dollars in the value of produce as $H^V is 
contained times in $ 6 3 2 9 ^ 8 -, or $600. 

(8) 2% of $600 = $12, commission. 

(9) $600 — $12 = $588, net proceeds. 

(10) $588 + $42 = $630, amount to invest in goods at 5 ct. 

commission. 

(11) Then, \H of $630 = $600, value of the goods. 

PROBLEM 266. 

An article cost $6; at what price must it be marked so that the 
marked price may be reduced 22% and still 30% be gained? 

Solution. 

(1) $6 = cost price. 

(2) 30% = gain. 

(3) 30% of $6 = $1.80. 

(4) $6 + $1.80 = $7.80, selling price. 

(5) 100% = marked price. 

(6) 22% = the deduction. 

(7) 100% — 22% = 78%, selling price. 

(8) 78% = $7.80. 

(9) 1% = J B of $7.80 = *-10. 

(10) 100% = 100 times $.10 = $10, the marked price. 



122 FAIRCHILD'S SOLUTION BOOK. 

PROBLEM 267. 
A boy was to receive 25 ct. per day for prompt attendance at school 
for a term of 80 days, but had to forfeit 75 ct. for eaeh day absent; he 
received $15: find the number of days present. {Putnam Co.) 

Solution. 

(1) Had he attended the 80 days he would have received 80 

times 25 ct., or $20. 

(2) He lost by his absence $20 — $15, or $5. 

(3) Each day absent he lost $.75 + $.25, or $1. 

(4) Hence, to lose $5 he must have been absent as many days 

as $1 is contained times in $5, which is 5 days. 

(5) .'. 80 days — 5 days = 75 days, present. 

PROBLEM 268. 
A merchant sold | of his flour at 12% profit, % at 10% profit, and £ 
at 8% loss: how should he sell the remainder so as to gain 5% on the 
whole? (/. B. Maurer.) 

Solution. 

(1) Let 100% = cost. 

(2) i of 100% = 50%, what he sold at 12% profit. 

(3) 12% of 50% = 6%; 50% + 6% = 56%, what he received 

for the 50%. 

(4) i of 100% = 25%, what he sold at 10% profit. 

(5) 10% of 25% = 2i%; 25% -f- 2{% = 27-|%, what he re- 

ceived for the 25%. 

(6) i of 100% = 12|%, what he sold at a loss of 8%. 

(7) 8% of 12|% = 1%; 12|% — 1% = ll£%, what he re- 

ceived for the 12i%. 

(8) 50% + 25% + 12^% = 87-i%, amount sold. 

(9) 100% — 87i% — 12|%, what remains to be sold. 

(10) 56% + 271% -f- Hi - 95%, vvhat he received for amount 

sold. 

(11) 100% + 5%) = 105%, what he asked for his flour, since 

he wanted to gain 5% on the whole. 

(12) 105% — 95% = 10%, what he had to receive for the 

balance. 

(13) In order to sell the remainder, 12^%, for 10%, he would 

sell it for 10% -^ 12£% = 80% of the cost, or at a loss 
of 20%. 

PROBLEM 269. 
Ralph sold cotton and invested the proceeds in stock at 20% dis- 
count, receiving- 4% commission in each transaction; sold this stock at 
4% premium, commission 5%: what did the cotton sell for, my whole 
commission being $1780? 



COUNTY EXAMINA TION LISTS. 123 

Solution. 

(1) Out of each $1 received for cotton, my agent received 4 

ct. commission; 96 ct. = the proceeds. 

(2) Now, this proceeds is made up of 100 parts investment 

and 4 parts commission. 

(3) 104 = the whole number of parts; \%$ of 96 ct. = 92 T ^ 

ct., investment in stock. 

(4) Since the stock was bought at 80 ct. and sold at 104 ct., 

the amount of sales on stock was (92 T 4 g ct. X 104) -r- 
80 = 120 ct., and the commission on this is 120 ct. X 
.05 = 6 ct. 

(5) $1 — 92 T \ ct. + 6 ct. = 13 T 9 3 ct., whole commission on 

$1 of the cotton. 

(6) Hence, there will be as many dollars of receipts for cot- 

ton as 13 T %ct. is contained times in $1780, or $13000. 

.-. The cotton sold for $13000. 

PROBLEM 270. 

A broker bought 6% bonds at 20% premium, and kept them 5 years, 
when they were redeemed at par: what rate of interest did he make on 
the investment? 

Solution. 

(1) Let $100 = the bond. 

(2) The interest on $100 for 5 years at 6% = $30. 

(3) Then, $120 = cost of the bond, and 

(4) $130 — $120 = $10, the gain. 

(5) $10 H- $120 = 8J%. 

(6) 8J% -r- 5 = If %. 

.'. He made lf%. 

PROBLEM 271. 

A sells a horse to B at a gain, and B to C at the same rate of gain 
for $16; if B had sold for $10, his loss would have been half what he 
now gains: find what A paid. 

Solution. 

(1) It is obvious that as often as he loses $2 he gains $4. 

(2) The difference between $16 and $10, or $6 = the loss + 

the gain. 

(3) $6 is 1-J- times B's gain; then, $6 -7- l-£ = $4, B's gain. 

(4) $16 — $4 = $12, B's cost. 

(5) 100% = B's cost. 

(6) $12 = 100%, and $1 = & of 100% = iff %. 



124 FAIRCHIL&S SOLUTION BOOK. 

(7) $4 = 4 times iff-% = 33£%, A and B's gain. 

(8) 100% = A's cost. 

(9) 133£% = A's selling price. 

(10) 133J% = $12, A's selling price. 

(11) 1% = $12 4- 133J = $.09. 

(12) 100% = 100 times $.09 = $9. 

.-. A paid $9. 

PROBLEM 272. 

A sold some cotton by his ag-ent, giving 4t% commission, and in- 
vested his proceeds in pike stock 20% below par. Having- waited for a 
favorable turn, he has, this morning-, sold his stock so as to g-ain upon 
it a sum equal to 14% of his cotton. Allowing- me a rate of commission 
just £ of the present discount of pike stock, he has advised me to invest 
for him in a note which will, in 8 mo., at 6% bring- $268.12^: how much 
will be his whole g-ain upon the value of his cotton? 

Solution. 

(1) 100% — 4% = amount invested in stock, the stock is 

worth 80%. 

(2) Now, if the gain on 96% is 14%, on 80% the gain is £§- 

of 14%, or 11|%. 

(3) Value of stock = 80% + llf % = 91f %. 

(4) 100% — 91| = 8J% discount. 

(5) f of 8J% = 6|%, my rate of commission. 

(6) Out of every $1.06J received for stock, $1 is invested in 

1061- 
the note, or ■-— = H of the note = amount received 

100 * 5 

for the stock. 

(7) The present worth of $268,121- for 8 mo. at 6% =$268.12*. 

which divided by 1.04 = $257.81^. 

(8) .-. The stock brought |-f of $257.81^ = $275. 

(9) It cost ^r 2 of $275 = $240. 

IT 

(10) Hence, the cotton was sold for $240 -± .96 = $250, and 
the whole gain is $268,121 — $250 = $18.12*. 

PROBLEM 273. 
Suppose 10% railroad stock is 20% better in market than 5% canal 
stock, how much money is invested in each if my income from each in- 
vestment is $600, and the whole investment pays 5%? 

Solution. 

(1) 10 ct. = income on $1 of railroad stock; $600 = total 
income on railroad stock. 



COUNTY EXAMINATION LISTS. 125 

(2) $600 -r- .10 = $6000, face value of railroad stock. 

(3) 5 ct. = income on $1 of canal stock; $600 = total in- 

come on canal stock. 

(4) $600 ~ .05 = $12000, face value of canal stock. 

(5) If railroad stock is 20%, or \ better in the market than 

railroad stock, then the cost of $lj of railroad stock is 
f of the cost of $1 of canal stock; and if there were 
just so many dollars in face value of railroad stock as 
there are in face value of canal stock, then its cost 
would be f of the cost of canal stock. 

(6) But a comparison of the two shows that there are only 

half as many dollars of railroad stock as there are of 
canal stock; hence, f of \ of canal stock = T \ of canal 
stock = cost of railroad stock expressed in terms of 
canal stock. 

(7) \% = cost of canal stock, and -J-J -\- -^ = -}-$, or cost of 

both. 

(8) $600, income on railroad stock + $600, income on canal 

stock = $1200, total income. 

(9) Each $1 of investment yields an income of 5 ct. 

(10) $1200 t- 5 ct., or $24000 = total cost, or investment. 

(11) .-. {$ = $24000. 

(12) t l = T V of $24000 = $1500. 

(13) T 6 o = 6 times $1500 = $9000, cost of railroad stock. 

(14) {% = 10 times $1500 = $15000, cost of canal stock. 

PROBLEM 274. 

If a man can make 5%, payable annually, on his money, what can 
he afford to pay for a 100-dollar 4% bond that is due in 2 years? 

(Putnam Co.) 
Solution. 

(1) Amount of $1 in 2 yr. at 5%, payable annually, is $1.1025. 

(2) The interest on the 100-dollar bond for 2 yr. at 4% is $8, 

and $108 is the amount of the bond. 

(3) Hence, he can pay $108 -7- 1.1025, or $97,969+ for the 

bond. 

PROBLEM 275. 

The difference between the true discount and the bank discount of 
a note due in 60 days, at 6%, is 24 cents: what is the face of the note? 

(Putnam Co.) 
Solution. 

(1) True discount on $1 is $1 — ($1 -7- 1.01) — $.00fj»f 

(2) The bank discount on $1 for the time and rate is $.01. 



126 FAIRCHILD'S SOLUTION BOOK. 

(3) Now, $01 — $.00}$} = $$.00 T -L-, the difference. 

(4) $24 -^ $.00^ = $2424, the face of the note. 

PROBLEM 276. 

A dealer in stock bought 1270 head of stock, and f of the number of 
horses is to f of the number of cows as f is to f- : find the number of 
each. {Putnam Co.) 

Solution. 

(1) | no. of horses : -| no. of cows :: f : f, or f^ no. of 

horses = |^ no. of cows. 

(2) 3*2 no. of horses = ^ T of f jj- = -^fa no. of cows. 

(3) ff, tne number of horses = 32 X ^fa — fff , no of cows. 

(Number of horses in terms of cows.) 

(4) |-j-g- = no. of cows, or there are 64 horses to 63 cows. 

(5) 64 + 63 = 127; T %\ of 1270 = 630 cows. 

(6) tW o f 12 ?0 = 640 horses. 

.*. He bought 630 cows and 640 horses. 

PROBLEM 277. 

If an article had cost me 20% more, my rate of g-ain would have 
been 25% less: find the rate of g-ain. {Putnam Co.) 

Solution. 

(1) The selling price divided by the cost gives the amount 

received for $1 of cost. 

(2) Since the selling price is the same in each case, the dif- 

ference between the amounts is the difference in the 
gain %. 

(3) Hence, -j-J-q — y-J^ = g-J^, the difference in the selling price. 

(4) eio = t 2 o 5 o» or i °f selling price is 25%, and | = 6 times 

25% = 150%?, the selling price. 

(5) .-. 150% — 100%, the cost = 50%, gain. 

PROBLEM 278. 

If my cost price had been 20% less, my rate of loss would have been 
15% less: what was my rate of loss? {Little Rock, Ark.) 

Solution. 

(1) Solution is the same as the above; hence, fa — -j-J^ = ^J-^, 

the difference in selling price. 

(2) totj — tVo» or i °f selling price is 15%?, and | - — 4 times 

15% = 60%, the selling price. 

(3) 100%, the cost — 60%, the selling price = 40%, loss. 



COUNTY EXAMINATION LISTS. 127 

PROBLEM 279. 
Bought rice at 6| cents a pound; waste by transportation and re- 
tailing- was 5% ; interest on first cost to time of sale was 2% : how much 
must be asked per pound to gain 20? {N. Y. list.) 

Solution. 

(1) Let 100 = number of pounds bought. 

(2) 100 X 6-1 ct. = $6.50, cost of the rice. 

(3) Interest at 2% = $.13; $6.50 + $.13 = $6.63, the whole 

cost. 

(4) 125% of $6.63 = $8.28f, amount he must charge. 

(5) 5% of 100 lb. = 5 lb., and 100 lb. — 5 lb. = 95 lb. 

(6) .-. $8.28f -7- 95 = 8f| cts. 

.'. I must ask 8f| ct. per pound. 

PROBLEM 280. 
When U. S. bonds are quoted in London at 108| and in Philadel- 
phia at 112£, exchange $4.89£, gold quoted at 107, how much more is a 
$1000 U. S. bond worth in London than in Philadelphia? (Putnam Co.) 

Solution. 

(1) $1000 X 1.124 — $1122.50, price in Philadelphia. 

(2) $1000 X 1.08f = $1087.50, price in London. 

(3) $1 of London gold is worth $1.07 of Philadelphia currency. 

.-. $1087.50 X 1.07 = $1163.621, price of London bond 
in U. S. currency. 

(4) $1163.62| — $1122.50 = $41.12^, what the London bond 

cost an American more than the Philadelphia bond. 

(5) To find the difference in cost to an Englishman in Lon- 

don, we proceed as follows: 

(6) $1000 x 1.12} = $1122.50. 

(7) $1122.50 -i- 1.07 = $1049.06^*7, price of the Philadelphia 

bond in English gold. 

(8) $1000 X 1.08f = $1087.50, and $1087.50 — $1049.06^ = 

$38.43,V 7 . 

(9) $38.43 t Vt -t- $4.89-l = £7 17s. .433d. 

(Prof. J. B. M. Zerr, Ph. D.) 

PROBLEM 281. 
A Texan farmer owns 5169 cattle; there are 3 times as many 
horses as cows, plus 569, and 4 times as many cows as sheep, minus 126: 
how many has he of each? (Brooks' 1 Arith.) 

Solution. 

(1) It is obvious that the number of cattle when there are 4 
times as many cows as sheep and 3 times as many 
horses as cows = 5169 -f 126 — 569 = 4726. 



128 FAIRCHIL&S SOLUTION BOOK. 

(2) Now, as often as he takes 1 sheep, he takes 4 cows and 12 

horses, or 17 in all. 

(3) .'. 4726 -f 17 = 278, no. of lots of 1 sheep, 4 cows, 12 

horses. 

(4) .'. 278 X 1 = 278, number of sheep. 

(5) 278 X 4 — 126 = 986, cows. 

(6) 278 X 12 + 569 = 3891, horses. 

PROBLEM 282. 

I owe A $100 due in 2 years, and $200 due in 4 years: when will the 
payment of $400 equitably discharge the debt, money being worth 3#? 

Solution 

(1) $100 on interest for 2 yr. at 6% = $12. 

(2) $200 on interest for 4 yr. at 6% = $48. 

(3) $400 on interest for 1 yr. at 3% = $12. 

(4) ($12 + $48) -=- $12, the interest on $400 for 1 yr. at 3% 

= 5 years. 

.'. 5 years is the required time. 

PROBLEM 283. 

Fritz owes James $200 due in 2 years, and James owes Fritz $100 
due in 4 years: when can Fritz pay James $100 to settle the account 
equitably, money being worth 6%? 

Solution. 

(1) The present value of $200 due in 2 yr. is $200 -i- 1.12 = 

$178,571. 

(2) The present value of $100 due in 4 yr. is $100 -r- 1.24 = 

$80,645. 

(3) $178 571 — $80,645 = $97,926, amount due James now. 

(4) $97,926 on interest at 6% yields $5,876 in a year, 

(5) $100 — $97,926 = $2,074, the interest which must accu- 

mulate in order that it may equal $100. 

(6) .'. $2,074 will accumulate in $2,074 -J- 5.876, or .3529 yr. 

PROBLEM 284. 

Mr. Merchant sells 20% above cost, with weights and measures 12£ 
% "short," allows a discount of $5 on every bill of $50, and loses 5 fa of 
his sales as "bad debts." Find his rate % of net profit or net loss, 1 
cent in every $1 of sales proving counterfeit, and collection charges 
being 2£%. {Mathematical Monthly.) 

Solution. 

(1) 100% — 12|% = 87|%, what he sells for 120%. 

(2) 120% rr- 87-£% = 137-1%, what he gets for 100%. 



COUNTY EXAMINA TION LISTS. 129 

(3) $5 on $50 is 10%. 

(4) 10% + 5% + 1% + 2|% =3 18^%, what he loses. 

(5) 1374% of 18*% - 25«%. 

(6) 137|%-25tt% = lllH. 

.-. He gains llf£%. (G. ^. ^. Z^r, 

PROBLEM 285. 
Bought an article and sold it for 3% less than it cost me; bought 
it back, paying- 3% more than I sold it for; I lost $12 by the transaction: 
what did the article cost at first? 

Solution. 

(1) 100% = the cost of the article. 

(2) 97% = the selling price. 

(3) 3% of 97% = 2.91%. 

(4) 3% + 2.91% = 5.91%, loss. 

(5) .-. 5.91% = $12. 

(6) 1% = ^.Vr of $12 = $2.0305. 

(7) 100% = 100 times $2.0305 = $203.05, cost of the article. 

PROBLEM 286. 
A sells goods which cost him $160, at a certain rate of gain; B sells 
to C at the same rate of gain; C pays $360 for the g-oods: find the rate 
of gain for A and B. 

Solution. 

(1) 100% - cost of goods. 

(2) r = rate of gain. 

(3) 100% + r = selling price. 

(4) 100% + r = 100% + r times $160 = $160 + 160r, A's 

selling price and also B's cost. 

(5) 100% = cost of goods by 2d condition. 

(6) 100% + r = selling price. 

(7) 100% + r = 100% + r times $160 + 160r = 160r a + 

320r -\- $160 = selling price, or C's cost. 

(8) 160r* + 320r + $160 = $360. 

(9) 4r* + 8r + $4 = $9. 

(10) r* + 2r = $|. 

(11) r = $.50, or50%. 

.*. A's and B's rate of gain is 50%. 

PROBLEM 287. 
A man left his property to three sons, to A £ wanting $180, to B £, 
and to C the rest, which was $590 less than A and B received: what was 
the estate? {Teachers' Review.) 



130 FAIRCHIL&S SOLUTION BOOK. 

Solution. 

(1) 100% = the estate. 

(2) \ of 100% = 33^%, A's,share + $180. 

(3) 33|% — $180 = A's share. 

(4) \ of 100% = 25%, B's share. 

(5) 33^% — $180 + 25%, or 58£% — $180 = A's and B's. 

(6) 100% — (58^% — $180) = 41J % + $180, C's money. 

(7) (58J% — $180) — (41f % + $180) = 16f % — $360, or 

what A and B have more than C = $590. 

(8) 16 J % = $950. 

(9) 1% = $950 -i- 16f = $57. 

(10) 100% = 100 times $57 = $5700. 

.'. The estate was worth $5700. 

PROBLEM 288. 

My loss on the sale of one article is 25%, and my gain on another is 
20% ; the difference in the cost of the two articles is $20, and my lose is 
equal to my gain: what was the price of each article? 

Solution. 

(1) 100% = cost of one article. 

(2) 100% + $20 = cost of the other. 

(3) 25% of 100% = 25%, loss on one article. 

(4) 20% of (100% + $20) = 20% + $4, gain on the other. 

(5) 25% = 20% + $4. 

(6) 5% = $4; 1% = $.80. 

(7) 100% = 100 times $.80 = $80, cost of one article. 

(8) 100% -f $20 = $100, cost of the other. 

PROBLEM 289. 

If 5 acres of grass, together with what grows on it during the time 
of grazing, keep 20 oxen 10 weeks, and 8 acres keep 29 oxen 16 weeks, 
how many oxen will 15 acres keep for 6 weeks? 

Solution. 

(1) Suppose each ox eats 100 lb. of grass in a week. 

(2) Then, (20 X 10 X 100 lb.) 4- 5 = 4000 lb., quantity of 

grass on 1 A. in 10 weeks. 

(3) (29 X 16 X 100 lb.) -5- 8 = 5800 lb., quantity of grass on 

1 A. in 16 weeks. 

(4) (5800 lb. — 4000 lb.) H- (16 — 10) = 300 lb., grown on 

1 A. in 1 week. 

(5) 300 lb. X 10 = 3000 lb., grown on 1 A. in 10 weeks. 

(6) 4000 lb. — 3000 lb. = 1000 lb., original quantity on 1 A. 



COUNTY EXAMINA TION LISTS. 131 

(7) 1000 lb. X 15 = 15000 lb., original quantity on 15 A. 

(8) 300 lb. X 15 X 6 = 27000 lb., grown on 15 A. in 6 weeks. 

(9) 15000 lb. + 27000 lb. = 42000 lb., amount to be eaten 

from 15 A. in 6 weeks. 

(10) 100 lb. X 6 = 600 lb., what 1 ox eats in 6 weeks. 

(11) 42000 H- 600 = 70. 

.*. 15 acres will keep 70 oxen 6 weeks. 

PROBLEM 290. 

A and B found a sum of money; A takes $1000 and r \ of the 
remainder for his half: how much did they find? 

Solution. 

(1) Let 100% represent the sum. 

(2) 100% — $1000 = the remainder. 

(3) A of (100% — $1000) = w% — &4-F- 

(4) $1000 + 2ff-% — $*flp. - 50%. 

(5) Then, $1300 + 200% — $2000 = 650%. 

(6) 450% = $11000. 

(7) 1% = -fa of $11000 = $24.4444|. 

(8) 100% = 100 times $24.4444f = $2444.44*. 

.*. They found $2444.44f 

PROBLEM 291. 

A man bought a farm for $3448.10, and agreed to pay principal and 
interest in 4 equal annual installments: how much was the annual 
payment, interest being 5%? [Putnam Co.) 

Solution. 

(1) Evidently the payment exceeded the interest due, or the 

principal would not have been reduced. 

(2) Now, for each $1 paid on the principal the 1st time, there 

was paid $1.05 the 2d time, $1.1025 the 3d time, and 
$1.157625 the 4th time. 

(3) .*. $4.310125 was paid in the 4 payments as often as $1 was 

paid in the 1st. 

(4) But $3448.10 was paid on the principal in the 4 payments. 

(5) Then, the amount paid on the principal the 1st time is 

$3448.10 -r- $4.310125, or $800. 

(6) The 1st interest is $3448.10 X 5%, or $172,405. 

(7) $800 + $172,405 = $972,405, the 1st payment. 

.*. The payments are each $972,405. 



132 FAIRCHIL&S SOL UTION BOOK. 

PROBLEM 292. 

An interest bearing note dated August 1, 1892, was discounted at 
90 days at 8% ; the face of the note was $750, and the proceeds $759,982: 
what was the date of discount? 

Solution. 

(1) 90 da. + 3 da. of grace = 93 da. 

(2) Interest on $1 for 93 da. at 8% = $.0206f. 

(3) $1 — $.0206| = $.9793^, proceeds of $1. 

(4) $759,982 -^- .9793^ = $776.0197, the amount of the note at 

the end of the 90 da. 

(5) $776.0197 — $750 = $26.0197, interest. 

(6) Interest on $750 for 1 yr. at 6% = $45. 

(7) $26.0197 -j: $45 = the time, 6 mo. 28 da. 

(8) 6 mo. 28 da. — 90 da. = 3 mo. 28 da. 

(9) Aug. 1, 1892 + 3 mo. 28 da. = Nov. 29, 1892. 

.*. The note was discounted Nov. 29, 1892. 
Note.— Solved by Prof. G. B. M. 2err for the A. M. Monthly. 

PROBLEM 293. 

A company of fifty persons engage dinner at a hotel, but before 
paying the bill 10 of the persons withdrew, by which each person's bill 
was increased 25 ct.: what was the bill? 

Solution. 

(1) If the expense of one is increased $.25, the expense of 50 

is 50 times 25 ct., or $12.50, which the 10 men should 
have paid. 

(2) One man should have paid T V of $12.50, or $1.25. 

(3) 50 — 10 or 40 times $1.25 = $50. 



The bill was $50. 



PROBLEM 294. 



Sixteen men hire a coach for a certain sum, but by taking in 4 more 
persons the expense of each is diminished $$-: what did they pay for 
the coach? 

Solution. 

(1) If the expense of one is diminished %\ , the expense of 

16 is diminished 16 times $J, which is $2, which the 4 
men pay. 

(2) If 4 men pay $2, one man pays ^ of $2, or $J. 

(3) 16 + 4, or 20 men, pay 20 times $|, which is $10. 

.*. They pay $10 for the coach. 



COUNTY EXAMINATION LISTS. 133 



BOAT PROBLEMS. 

PROBLEM 295. 

A vessel sails 15 miles an hour in still water; going up stream it re- 
quires 72 minutes to sail 15 miles: how long will it require to go down 
stream 15 miles? 

Solution. 

(1) Rate up stream is 15 -r- f hr. = 12-|- mi. an hr. 

(2) The rate of current is 15 mi. — 12\ mi. = 2\ mi. per hr. 

(3) Then, the rate down stream is 15 mi. -f- 2\ mi., or 17£ mi. 

an hour, and it will take 15 -r- 17^ X 60 = 51f min. 

.*. The vessel goes 15 mi. down stream in 51-f min. 

PROBLEM 296. 

I can row up stream in 64 minutes and back again in 60 minutes: 
determine the distance, the rate of current being \ mile per hour. 

Solution. 

(1) In 1 minute I can row up stream 6 l ? of the distance, and 

Jjs of the distance back again in the same time. 

(2) Hence, twice the distance the current moves in a minute 

is ^ — ^ T = - 9 -| of the distance. 

(3) It is true that the rate of the current will be half the dif- 

ference of the rates of sailing down and up; that is, ^ 
°f ¥ftt» or oVtf °f tne distance the current moves in 1 
minute. 

(4) The current in 1 hour moves r9 V X 60 = xff-o, or zh °f 

the distance, or -\ a mile. 

(5) $-$, the distance = 32 X \ of a mile, or 16 miles. 

.*. The distance is 16 miles. 

PROBLEM 297. 

If a boat goes down stream 42 miles and back in 20 hours, going 14 
miles down in the same time that it goes 6 miles up, what is the rate in 
still water, and what is the rate of current? {The Teachers' Review.) 

Solution. 

(1) The time down stream is to the time up stream as ^isto 

£, or as 6 to 14. 

(2) Hence, it takes ^\ of 20 hr., or 6 hr., to go down and ^ of 

20 hr., or 14 hr., to come up. 

(3) The rate down stream is 42 -f- 6 = 7 mi. an hr., and the 

rate up, 42 H- 14 = 3 mi. an hour. 



134 FAIRCHILD'S SOLUTION BOOK. 

(4) The rate of current must be (7 — 3) -r- 2 = 2 mi. an hr., 
and the rate of sailing in still water is 7 — 2 = 5 mi. 
per hr. 

PROBLEM 298. 

Frank would row a boat 7 miles with the tide in the same time he 
would row 5 miles against it; but if the current ran half a mile an hottr 
more, he would row twice as rapidly with the tide as against it: find 
his rate in miles per hour in still water. 

Solution. 

(1) 7 mi. = rate per hr. with the tide. 

(2) 5 mi. == rate of rowing against the tide. 

(3) (7 mi. — 5 mi.) -=-2 = 1 mi., rate of tide. 

(4) 7 mi. — 1 mi. = 6 mi., the rate of rowing. 

(5) 1 mi. -=- 6 mi. = \\ :. the rate of the tide is \ of the 

rowing rate. 

(6) Then, by the second condition he rows 2 mi. with the 

tide and 1 mi. against it. 

(7) (2 mi. — 1 mi.) -r- 2 = \ mi., rate of tide. 

(8) 2 mi. — \ mi. = 1^ mi., rate of rowing. 

(9) \ mi. -s- \\ mi. = \\ :. the rate of tide is \ of the rowing 

rate. 

(10) Hence, the additional \ mile per hr. is \ — £, or \ of 

rowing rate. 

(11) .*. \ = \ mi.; f, the rowing rate = 6 times \ = 3 mi. 

.-. The rate of rowing is 3 mi. per hr. 



CHAPTER XII. 



MENSURATION. 



79. Mensuration is that branch of applied geometry which 
gives the rules for finding the lengths of lines, the areas of sur- 
faces, and the volumes of solids. 

80. A position in space that is without magnitude is called 
a Point. 

81. A Line is the path of a point in motion. Lines are 
represented upon paper or blackboard by marks made by 
pen or crayon. The point of the pen or crayon leaves a visi- 
ble mark. This mark has breadth and occupies some of the 
surface upon which it is drawn, and is called a physical line. 
If we continually diminish the breadth of the physical line we 
make it approximate to the geometric line. 

82. A Straight Line is a line traced by a point which does 
not change its direction of motion. 

83. A Curved Line is the path of a point which constantly 
changes its direction of motion. 

84. An Angle is the opening between two lines which meet 
each other. The point in which the lines meet is called the 
Vertex, and the lines are called Sides. 

85. A Right Angle (a square corner) is formed by one line 
standing on another so as not to incline on either side. 

86. An Acute Angle (a sharp corner) is an angle which is 
less than a right angle. 

87. An Obtuse Angle (a blunt corner) is an angle greater 
than a right angle. 

88. A Triangle is the figure formed by three lines and the 
determined points, or by three points and the determined 
lines. 

(135) 



136 



FAIRCHIL&S SOLUTION BOOK. 



89. A Scalene Triangle is one which has no two of its sides 
equal. 

90. An Isosceles Triangle is one which has two of its sides 
equal. 

91. An Eqnilateral Triangle is one whose sides are ail equal. 

92. A Right Triangle is one which has a right angle. The 
side opposite the right angle is called the Hypothenuse. 



I. TRIANGLES. 



PROBLEM 299. 

Prove that the square described on the hypothenuse of a right- 
angled triangle is equivalent to the sum of the other two squares de- 
scribed on the other two sides. 

Solution. 

LetABCbearight 
triangle. Produce BC 
to D, making CD = 
AB, and complete 
the square BDGF. 
Now make FH and 
GE = AB and com- 
plete ACEH. 

The triangles ABC, 
CDE, EGH and 
HFA are equal, and 
ACEH is the square 

on AC, or on the hypothenuse. On MN = BD construct the 
square MNPK. Arrange the four triangles of the first square 
as drawn in the second, OP is the square on AB, and OM on 
BC. From each of the two squares take the four equal tri- 
angles, and there remains the square ACEH, equivalent to 
the sum of the squares OP and OM. Q. E. D. 

NOTE. — The author gives this demonstration in class work, for it 
seems the simplest. 

The following is Garfield's demonstration of the right tri- 
angle: 




FIG. 



MENSURA TION. 



137 




FIG. 4. 

(AB X AC) 
(AB + AC) 5 



Let ABC be the given right-angled tri- 
angle. Produce AC to E, making CE = 
AB. At E draw ED perpendicular to CE 
and make it equal to AC. Draw DC and 
BD; then CD = CB and A BCD = ^BC 2 . 
TheareaofABDEisequalto|AE(AB+ED) 
or |( AB + A C) 2 ; it is also equal to 
|(BCxCD) + twice the A ABC or |BC 2 + 
(AB X AC). 

Therefore, K AB + AC) 2 = 4rBC 2 + 
whence, 

= |BC 2 + 2 (ABX AC), or AB 2 + AC 2 = BC 2 . 

O. E. D. 



PROBLEM 300. 



Given the right-angled triangle ABC, the base AC 
titude BC = <>: what is the hypothenuse? 

Solution. 

(1) AC = 40, the base. 

(2) BC = 9, the altitude. 

(3) 40 2 = 1600, the square of the 

base, AC. 

(4) 9 2 = 81, the square of the alti- 

tude, BC. 

(5) 1 600+ 81 = 1681, AC 2 + BC 2 . 

(6) V1681 = 41, AB. 



40 and the al- 




PIG. 



. 41 = the hypothenuse. 

PROBLEM 301. 

The hypothenuse of a right-angled triangle is 97, and the altitude 
05: what is the base? 

Solution. 

(1) AB = 97, the hypothenuse. 

(2) BC = 65, the altitude. 

(3) 97 2 = 9409, the square of the hypothenuse AB. 

(4) 65 2 = 4225, the square of the altitude BC. 

(5) 9409 — 4225 = 5184, the difference of the squares of the 

hypothenuse and altitude. 

(6) V5184 = 72 = AC. 
.'. 72 = the base." 

PROBLEM 302. 

What is the area of a right triang-le whose base is 24 feet and alti- 
tude 7 feet? 



138 



FAIRCHILD'S SOLUTION BOOK. 




Solution. 

(1) We can see at once that 

ACB = AEB. 

(2) AC = 24fi., the base. 

(3) BC = 7 ft., the altitude. 

(4) 24 X 7 = 168 sq. ft., the 

product of the base and 
altitude. 

(5) \ of 168 sq. ft. = 84 sq. ft., 

half the product of the base and altit 

.'. 84 sq. ft. = the area of ABC. 

PROBLEM 303. 

What leng-th of rope will be required to draw water from a well, it 

being- 38 feet to the water, the sweep to be supported by an upright 

post 20 feet high, and standing- 20 feet from the well, and the foot of 

the sweep to strike the ground 20 feet from the foot of the upright post? 

Solution. 

(1) AE = 20 ft.; FE, the post = 

20 ft.; BE = 20 ft.; and 
CB will be the required 
length of^the rope. 

(2) By the similar triangles ABC 

and AEF, we have AE : 
EF :: AB : CB, or 20 ft. 
: 20 ft. :: 40 ft. : (CB = 
40 ft.) 

Note. — This solution was prepared by the author for the American 
Mathematical Monthly. 

PROBLEM 304. 
The hypothenuse of a right-angled triangle is 97 ft., and the per- 
pendicular is 65 ft.: required the base, without squaring either of the 
given numbers. 

Solution. 
[ (97 + 65) (97 — 65) ] = 72 ft., the base. 

PROBLEM 305. 
How long a ladder will be required to reach a window 40 feet from 
the ground, if the distance of the foot of the ladder from the wall is 
one-fifth of the length of the ladder? 

Solution. 

(1) From Fig. 5, AB, or 5*, represents the length of the lad- 

der, and AC, or x, the distance the foot of the ladder is 
from the wall. 

(2) CB = 40 ft. 




FIG. 7. 



MENSURA TWN. 



139 



(3) Then we have 25* 2 + x* — 40 2 . 

(4) Whence, 5* = 40.824829 ft., the required length. 

Note. — This solution was prepared by the author for the School 
Visitor. 

PROBLEM 306. 

A right-angled triangle, altitude 7 ft., and hypothenuse 25 ft., has 
the same area as a rectangle whose sides are as 7 to 3: find dimensions 
of the rectangle. 

Solution. 



7 2 ) 



24 ft. 



(1) The base of the triangle is V (25 2 

(2) Its area is 7 X 24 -f- 2 = 84 sq. ft. 

(3) Assuming a rectangle 7 ft. long and 3 ft. wide which is sim- 

ilar to the given rectangle, we fino its area to be 21 sq. ft. 

(4) The areas of similar figures are to each other as the 

squares of their homologous sides; then 21 sq ft. : 84 
sq. ft :: 7 2 : (the required length = 14 ft.) 

(5) 21 sq. ft. : 84 sq. ft. :: 3 2 : (the required width = 6 ft.) 

.*. The dimensions of the rectangle are 14 ft. and 6 ft. 

PROBLEM 307. 

Two trees, the first of which is 60, and the second 100 feet high, 
are 300 feet apart: what is the length of a string which, from a point 
between the trees, will just reach the top of each? 

Solution. 

(1) AD = 60 ft., the height of 

the tree AD, and BC = 
100 ft., the height of the 
tree BC. 

(2) Connect the tops of the 

trees by the line DC, 
and from the middle 
point T of DC let fall 
TF perpendicular to AB. 
Draw TE perpendicular 
to DC. 

(3) CS = 100 — 60, or 40 ft. 

(4) TF = |(BC + AD) = 80 ft. 

(5) By similar triangles, DS : SC :: TF 

40 ft. :: 80 ft. : (FE = 10} ft.) 

(6) EB = FB — FE = 150 — 10} = 139^ ft 

(7) EC = V(EB 2 + BC 2 ) - 171.50+ ft. 
.-. CE or DE = 171.50 ft. 




FIG. 8. 



FE, or 300 ft. 



140 



FAIR CHILD'S SOLUTION BOOK, 



PROBLEM 308. 

A ladder 61 ft. long being- placed 7 ft. from the center of a street 
will just reach the top of a tower on one side and 6 ft. higher on the 
other side: find width of street. 

Solution. 

(1) From Fig. 8, AD = x — 3 and BC = x + 3. 

(2) AB = 2y, the width of street. 

(3) Then, from the two right triangles AED and EBC, 

+ ?) 2 + {y — 3) 2 = 61 2 , and (y — 7) 2 + (* + 3) 2 
= 61 2 . 

(4) Expanding and subtracting one from the other, 28j = 12*. 

(5) .:'x = %y. 

(6) Subtracting in either equation ^fy> 2 — 3663. 

(7) y — 23.841 ft., 2y = 47.682 ft., and AD, or x — 3 = 

52.62 ft., BC, or x + 3 = 58.62 ft. 



PROBLEM 309. 

There are two pillars, one 110 ft. high and the other 120 ft. A lad- 
der placed between the pillars will reach the top of either without mov- 
ing it at the base, and it is further found that the ladder will reach 
from the top of one pillar to the top of the other: determine the length 
of ladder and distance between the poles. 

Solution. 

(1) Let AB = width of street. 

(2) AF = 110 ft., BC = 120 ft., 

and CE = 10 ft. 
Let x = FP = PC = CF, 
the length of the ladder. 

FE 



(3) 
(4) 



AB = V(*' 



10 2 ) : 



AP = V(* 2 — HO 2 ), and 
PB = V(* 2 " 120 2 ). 



(5) V(-* 2 — 10 2 ) =V0 2 — no 2 ) 

+ VO* 2 — 120 2 ). 

(6) Squaring and reducing, x 2 = £*£& 

133.1665 ft 




FIG. 9. 



(7) x 



BP 



= V(^ 2 — no 2 ) = 

.-. x = 133.1665 ft., PA 



VC* 2 — 120* ) 
57.735 ft. 



dt .( 



35 ft. PA 



75.0555 ft., and PB = 57.735 ft. 



PROBLEM 310. 

There are two buildings, the heights of which are 40 and 29 feet, 
and a ladder is placed against the higher building in such a manner 



MENSURA TION. 



141 



that a line 46 feet long-, reaching- from the top of the lower to the mid- 
dle of the ladder, is the least that will reach it from there to any part 
of it: required, the length of ladder, and the distance its foot is from 
each building. ( Whitney.) 

Solution. 

(1) AF = 29 ft., height of the 

lower building. 

(2) BD = 40 ft., height of the 

higher building. 

(3) BE = x % the distance be- 

tween the foot of the lad- 
der and the higher build- 
ing,/ = the distance AE, 
and 2z = ED, the length 
of the ladder; FS = 46 ft. 

(4) Draw FD and FE; then EDF is an isosceles triangle. 




(1) 
(2) 
(3) 
(4) 
(5) 



(5) Then we have x* + 1600 = 4s-' 

(6) {x+y)* + (40 — 29)* = y* _|_ 29 2 

(7) j 2 + 29* = ^ 2 + 46 2 . . . 

(8) Reducing (2), x* + 2xy = 720 

(9) From (3), z* = j>* — 1275 . 

(10) Substituting the value of z in (1), 2y = V (x* — 6700). 

(11) Substituting this value of 2y in (4), transposing, squar- 

ing and reducing, 407* = 25920; also x = 7.97* ft., v = 
41.12 ft., and 2z = 40.788 ft. 

PROBLEM 311. 

A tree 125 ft. high stands on the bank of a river 105ft. wide; where 
must the tree break off so that it may remain connected at the point of 
breaking - and its top just reach the opposite shore? 

First Solution. 

(1) Let AB = the height 

or sum of the altitude 
and hypothenuse. 

(2) AG = the hypothe- 

nuse, and GB = the 
altitude. 

(3) Complete the square 

ABFL; its area will 
be 125* ft., or 15625 
sq. ft. 

(4) Let BC = GD; draw 

GK parallel to BF 
and NC parallel to 
AB. FIGp n> 




142 FAIRCHILD'S SOLUTION BOOK. 

(5) GC = the square of the altitude, and DL = the square 

of the hypothenuse. 

(6) Take LH equal to the base of the triangle formed by 

the broken tree, and complete the square SL. 

(7) Then, SL is the square of the base, which we take from 

the square of DL; the difference is the gnomon. 

(8) hh == the square of the altitude GC. 

(9) But if from ABFL, or 15625 sq. ft., we takeSL, or 11025 

sq. ft., the remainder is GC + DF and AD -f AM (AM 
= hh) = 4600 sq. ft. 

(10) Then we find \ of 4600 sq. ft., or 2300 sq. ft. = GC + CK 

or AD + AM. 

(11) But BC + CF = 125 ft. 

(12) Then 2300 sq. ft. ^- 125 = 18| ft., the altitude, or BC. 

.'. The tree must break off 18$ ft. from the ground. 

Note. — This solution was prepared by the author for the School 
Visitor. 

Second Solution. 

(1) Let AB = b, the width of the stream, 

and draw BD perpendicular to it = 
the height of the tree. 

(2) Join AD, and bisect it in E by the per- 

pendicular EC; join AC. 

(3) It is evident that AC = CD and AC + 

CB = BD = a, or the sum of the hy- 
pothenuse and perpendicular. 

(4) If x — the perpendicular, a — x will be 

the hypothenuse AC. 

(5) (a — x) 2 = b 2 + x\ or a* + 2ax + x 2 ^ Q n 

= b 2 — x 2 . 

a 2 fr2 

(6) Whence x = - — ^ . Now substitute the values of a 2 , 

b 2 and a, and x == 18| ft. 

Utile. — From the square of the height subtract the square of the 
base, and divide the difference by twice the height. 

PROBLEM 312. 

Find the area of a circle inscribed in a right-angled triangle whose 
legs measure 15 and 20 ft., respectively. 




MENSURA riON. 



143 




Solution. 

(1) Let AC = 20 ft., the base of 

the triangle ABC, and BC 
= 15 ft., the altitude. 

(2) AB, the hypothenuse = 

V(AC 2 -h~BC 2 ), or 25 ft. 

(3) FDE = the inscribed circle; 

OD, OF, OE = the radii fig. is. 

of the inscribed circle. 

(4) ^(AC X BC) = 150 sq. ft., the area of the AACB. 

(5) The radius of an inscribed circle is found by dividing 

twice the area of the triangle by its perimeter, or (150 
X 2) -^ 60 = 5 ft., the radius OE or OD. 

(6) Hence, 5 2 X n = 78.53 sq. ft. 

.*. 78.53 sq. ft. is the required area. 



PROBLEM 313. 

A ladder stands against the side of a house; if it is pulled out at the 
bottom 16 ft., it will not reach the top by 4 ft., and it is long enough 
to reach the top: find the length of the ladder. 



Solution. 



(1) Let DC = the 
height of the 
house, and AB 
= the ladder 
drawn out 16 ft. 
on CB. 

(2) DC = AB. 

(3) Construct t h e 
square DF up- 
on DC = the 
square on AB. 

(4) The square SF 
= the square on 
AC, the perpen- 
dicular of the 
A CBA. 

(5) 




FIG. 14. 



Let b = AD, or 4 ft. = the distance 
down when it was pulled out 16 ft. 

(6) aa' = the area of the rectangles ES and SC 

the area of the square DS. 

(7) .-. FO — FG 2 = aa'b\ or 16* = 256 sq. ft. 



the ladder slipped 
and b 2 — 



144 



FAIRCHILD'S SOLUTION BOOK. 



(8) aa'b' 1 — b 2 = 240 sq. ft., the area of the two rectangles 

ES and SC. 

(9) .-. The area of SC = | of 240 sq. ft. = 120 sq. ft. 

(10) SA = AD, or 4 ft.; then AC = \ of 120 sq. ft., or 30 ft. 

(11) AC -4- AD = 34 *ft. = DC, or AB. 

.*. 34 ft. is the required length of the ladder. 



PROBLEM 314. 

Two lines are drawn from the acute angles of a right triangle to 
the middle points of the opposite sides, and measure respectively, Vf3 
and V-^ : find the base and perpendicular. 

Solution . 

(1) Let ABC =the triangle. 

(2) EC = V52 and AT = 

V73. 

(3) Let repres e n t the 

square on the half of 
BC and ED half the 
square on AB. 

(4) Also the squares AX, 

PX and XK are each 
equal to ED. 

(5) Then, if we subtract 

from either of the 
squares ED or XK 
we have the gnomon 
S. 

(6) As EBC is a right tri- 

angle, 50+ S = 52 (1). 

(7) As ABT is also a right triangle, 50 + 4S = 73 (2), 

(8) Subtract (1) from (2) and solve for S, we find S == 

(9) Substitute the value of S in (1) we find that = 9. 

(10) /. BT=V9 = 3, or BC = 6. EB 
AB = 




FIG. 15. 



3. AC = VBC 2 + AB 2 
. ABC is a 6, 8 and 10 triangle. 



V* + S = 4, 

10. 



or 



PROBLEM 315. 

There are two ladders standing-] in the middle of a street; one is 
80 ft., and the other 90 ft., both lean ag-ainst a building- and the 90 ft. 
ladder reaches the top 20 ft. hig-her than the 80 ft. ladder: find heig-ht 
of the building. 



MENSURA T/ON. 



145 




FKr. V, 



Solution. 

(1) Let AC and DC be the 

position of the lad- 
ders. 

(2) BD* = the square LG, 

and BO — the square 
BM. 

(3) Subtract the square BM 

from the square AG 
= the two rectangles 
FL and LD plus the 
square AL = the 
square BM, = AC 2 — 
DC 2 , or 1700 sq. ft. 

(4) AD 2 = 400 sq. ft., the 

square AL. 

(5) 1700 sq. ft. — 400 sq. ft. == 1800 sq. ft., area of the two rec- 

tangles FL and LD. 

(6) LB or FL = 1300 sq. ft -r- 2, or 650 sq. ft.; as LD or AD 

= 20 ft., then DB = 650 sq. ft. -f- 20, or 8.25 ft. 

(7) AB = AD + DB, or 20 ft. + 32.5 ft. = 52.5 ft. 

.-. 52.5 ft. = the height of the building. 
PROBLEM :urt. 

A ladder 65 ft. long- is standing against a perpendicular wall; if 
the base were drawn out 02 ft., what distance would the top slide down 
the wall? 

Solution. 

(1) From Fig. 14, DC = AB = 65 ft. 

(2) DA = distance the top slipped down the wall, and CB = 
_52 ft. 

(3) CB 2 = aa'b*. 

(4) DC* — CB 2 , or aa'b'* — SF* = 1521 sq. ft. 

(5) Then, SG or AC = V1521 sq. ft. = 39 ft. 

(6) DA = DC — AD = 26 ft. 

.'. 26 ft. is the required distance. 

Notk. Such problems as the above and also Fig. 11 have bothered 
many an applicant at county examination-. But you see they are eas- 
ily sol red. 

PROBLEM 317. 
Inscribe the greatest possible rectangle in the right triangle whose 

sides are KM), HO. and f>0 ft. 




146 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) Let x = the length of the rectangle, 

and b the base, or 80 ft. 

(2) CA = a, AB = b, and FE — y\ then 

EB = b — x, xy =■ area of the rec- 
tangle = A maximum. 

(3) Differentiating, xdy + ydx — (1). ,. 

(4) The triangles FEB and ABC are 

similar; hence a : b :: y : $ — x\ ox ab — ax — by {2) 

(5) Differentiating, — adx = &£j/, whence d£j/ = ( — dwfo) -r £. 

(6) Substituting this value in (1), — axdx -r- b -\- ydx = 0. 

(7) Clearing, — «*^r + bydx = 0, and — ax + £j/ = 0. 

(8) by = ax . . . (3). 

(9) Then from (2) by = ab — ax. 

(10) .'. ab — ax = #*, ab = 2#*, £ = 2*, or * = £ -f- 2 and jj 

(11) Now as b = 80 ft., and a — 60 ft., by the above reason- 

ing, x = 40 ft., and y — 30 ft. 
.'. 40 ft. is the length, and 30 ft. the width of the maxi- 
mum rectangle. 

Note. — The sides of a maximum rectangle inscribed in a riglit 
triangle are respectively halves of the given leg's. 

PROBLEM 318. 

A pole 195 ft. higli, standing- on the margin of a pond which is 
108 ft. wide, was broken by the wind, the broken piece remaining - at- 
tached to the stump, while its extremity rested on a post in the pond, 
the post being 45 ft. above the surface of the lake; and while in that 
position the broken piece pointed exactly to the opposite edge of the 
pond: required the height of the stump. 

Solution. 

(1) From Fig. 17, AB == 408 ft., the width of the pond; FE 

= 45 ft., the height of the post. 

(2) x — BE, y = CD. AE = 408 — x, CF, or the broken 

part, is 450 — y; then x : 45 :: 408 — x : y\ then 
xy + 45* = 18360 ... (1). 

(3) Also, (408 — *)■ + r 2 = (450 —y) 2 , whence, * 2 — 816* 

+ 900j/ = 36036 . . . (2). 

(4) Also, 45 -r- x = (408 — x) +y; then * 2 —408* = 45jr (3). 

(5) From these three equations we find * = 108 ft., and 

y = 125 ft., and AC + 125 ft. + 45 ft. = 170 ft. 
.*. 170 ft. is the required height of stump. 

PROBLEM 319. 
The hypothenuse of a right-angled triangle is 35, the side of the 
inscribed square 12: what are the sides of the triangle? ( The Voice.) 



MENSURA TION. 



147 



Solution. 

(1) From Fig. 17, CB = 35, DF and FE = 12 ft. each, EB = 

x t AB = x + 12 and CD = y. 

(2) Then CA = y + 12. 

(3) Whence (* + 12)* + (j -f 12) 2 = 35 a , or x* + y* + 

24(x+y) =937 .. . (1). 

(4) From the similar triangles CDF and FEB, CD : DF :: 

FE : EB, or j : 12 :: 12 : x, from which xy = 144 (2). 

(5) Multiply (2) by 2, or by doubling (2) and adding (1), 

we then have y* + 2xy + x 2 + 24 (x + r) = 1225, or 
(x+y)* + 24(x+y) = 1225. 

(6) Completing the square, x ■ + y = 25 . . . (3). 

144 

(7) ;r = — from (2), and substitute its value in (3), whence, 



)" 



25 r = —144. 



(8) y — 12| = ±3i t or j = 16 or 9, and x = 9 or 16. 

.'. The base is 21 ft., or 28 ft., and the perpendicular is 
28 ft., or 21 ft. 



PROBLEM 320. 

The perimeter of a rig-ht triangle is 24; the hypothenuse is 2 longer 
than the altitude, and the altitude 2 longer than the base: what are its 
sides? 

Solution. 

(1) Let ABC be the triangle, 

and AG the square of 
the hypothenuse, AS 
the square of the alti- 
tude, and AK the .square 
of the base. 

(2) The rectangles nn and 

AX, etc., are equal. 

(3) The two rectangles nn' 

and the three squares 
aa'a" = the square on 
the base. 

(4) The square of the base 

— nn! = aa'a" = k, or 
any of the rectangles, or 12; then the length of the rect- 
angles = the length of the base. 

(5) Also the area of one of the rectangles k' is 12, and one 

side is four less than the other side. 




FIG. 18. 



148 



FAIRCHILD'S SOLUTION BOOK. 



(6) Then by Fig. 19, let WZ be 

a straight line equal to the 
difference of the sides of 
the rectangle k'. 

(7) Upon WZ as a diameter de- 

scribe a circle, and at the 
extremity of the diameter 
draw the tangent WF equal 
to the side of a square hav- 
ing the same area as the 
rectangle h'. 

(8) Through the point F and 

the center O, draw the se- 
cant FQ; then will FY, 
FQ be the adjacent sides 
of the rectangle. 

WF = V12 = WZ = 4, and WO 

FO 




FIG. 19. 



(9) 

(10) 
(11) 
(12) 
(13) 
(14) 



2. 



V(FW 2 + WO 2 ) = 4. 
FQ = FO + OO = 6, the base of the right triangle. 
AF = BC, or 6;~FD and DC each = 2. 
/. AC = AF + FD -f BC = 10. 
AB = AF -f FD, or 8. 
. ABC is a 6, 8 and 10 triangle. 



PROBLEM 321. 
A tree is standing one hundred feet from the bank of a run; a 
squirrel at the heig-ht of 60 ft. runs down and runs to the water's edge; 
at the same time a bird flies up and then down to the squirrel at the 
water's edge, and the bird flew as far as the squirrel ran: find the 
height that the bird flew. 

Solution. 

(1) Let E represent the 

point the bird is sit- 
uated when it flies 
upward and the 
squirrel runs down- 
ward. 

(2) Then x represents the 

height the bird flew, 
or CE. 

(3) BC =>, the hypothe- 

nuse of the right tri- 
angle ABC, whose 
base AB is 100 ft., 
and the perpendicu- 
lar AC = x + 60 ft. 

(4) Then (x + 60 ) 2 + fig. 20. 




MENSURA TION. 149 

100 2 =j/ 2 . . . (1). 

(5) x+y = 160 . . . (2). 

(6) Find the value oi y in (2), square, and substitute its value 

in (1), and solve for x. 

(7) Hence, x = 27/ T ft. 

.-, 27y\ ft. = height that the bird flew. 

PROBLEM 322. 

Two poles, perpendicular to the same plane, are 30 ft. and 50 ft. in 
height: at what height from the plane will lines drawn from the top of 
each to the base of the other, cross? 

Solution. 

(1) Let AC = 50 =^; DB = 

30 = b; BC = c, FE = 
x\ EB = y; CE = c — y. 

(2) Then by similar triangles, 

a : c :: x : y\ therefore, 
ay = ex . . . (1). 

(3) Also, /; : cw.x : c — x\ there- 

fore, be — by— ex . . (2). 

(4) Now from (1) and (2), 

ay-\-by — be, and y—bc-^r 

v . FIG. 21. 

(5) By substituting the value 

of y in the first proportion, we have a'.ew x : be -r- 
(a+b). 

(6) ex — abc- 1 r{a J rb), and x=ab-rr{a-\-b), or . 

(7) Hence, (50X30)^(50+30) =18$ ft. 

.\ 18f ft. is the required height, or FE. 

Exile. — Divide the pro duet of their heights by their sum. 

NoTK. — It would be well to bear in mind that the height at which 
the lines cross is not affected by the distance the poles are apart. 

PROBLEM 323. 

Two trees are a certain distance apart; from the top of a to the foot 
of t> is 60 ft.; from the top of b to the foot of a is 40 ft.; from the point 
where these two lines intersect to the plane is 15 ft.: what is the height 
of each tree and the distance apart? {American Mathematical Monthly.) 




150 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) From Fig. 21, let DB=*— y\ AC = x+y; FE = r = 15; CD 

= £ = 40; AB=A = 60; BF=^; and CE = w. 

(2) By similar triangles, w : c :: ze/-j-£ : x — y, or ^(^ — y) = 

c{zv+z) ... (1). 

(3) z \ c :: w-|-,s : x-\-y, or {w-\-£)c=.z(x-\-y) ... (2). 

(4) r.w{x-y)=z{x+y) . . . (3). 

(5) Put ;t:-f-j/=5 and x—y=.t. 

(6) Then wt=zs, or z#=— — . . . (4). 

(7) Substituting this value of w in the proportion of (2), 

2s zt-\-zs „. . .. . , z 

— \ c '.: - — - — : t. Dividing antecedents by — we 

it i 

have, s : c :: t-\-s : A or ^=<r(/-j-j). 

/nx „ T , 5/ (x-\-y)(x — y) x 2 — y 2 ._. 

(8) Whence, f=— = v T/V; ^ = Q J . . . (o). 

v 5-K x-\-y-\-x—y 2x v 

(9) From (5), # 2 — 2o:=j 2 , whence .*=<:+ VO^^ 8 )- 



(10) From figure, AB 2 — AC 2 =CB 2 , and CD 2 — DB 2 =CB 2 . 



(11) .-. AB 2 — AC 2 =CD 2 — DB 2 ,or« 2 — {x+y) 2 ~b 2 — (x— y)\ 
a 2 —b 2 



(12) Whence, x—- 



4y 

a 2 —b 2 



(13) Equating the two values of x, — j = ^-|"V (7 2 4~^ 2 )- 

(14) From this we have y*+\c(a 2 — b 2 )y = {a 2 — b 2 ) 2 . 

(15) Restoring numbers and solving the resulting equation, 

^2 A 2 fSOO 

i/ = 14.060811 ft. x=- — =^=35.620636 ft. x—y 

y y 



=21.499014 ft. AB = V(AB 2 — AC 2 ) = V[^ 2 — (^+^) 2 ] 
= 33.731178 ft. 

PROBLEM 324. 

An iron bar 20 ft. long- stands close against a vertical wall. If the 
lower end is drawn out and the upper slides down the wall until the 
bar assumes a horizontal position, what curve will a fly describe that 
remains upon the center of the bar, and what distance will it move? 



MENSURA TION. 



151 



Solution. 



(1) Let AB = the vertical wall, 

BC=the horizontal po- 
sition of the bar. 

(2) We see from the figure the 

different positions ofthe 
bar as it is drawn out. 

(3) The center of the hypothe- 

nuse or bar always main- 
tains the same distance 
from the right angle or 
foot of the wall; viz., \ 
the length of the ladder. 

(4) The middle point, there- 

fore, moves through a 
quadrant of a circle 
DKF, whose radius is IOX^tt 




FIG. 22. 



15.70795 ft. 



.*. The fly moves 15.70795 ft. 



PROBLEM 325. 

Given the area, a = 600, and the hypothenuse h = f>0: to find the 
other sides of the right-angled triangle by geometrical construction. 

First Solution. 



(1) Let ABC be the triangle; 

from B draw BD per- 
pendicular to the 
hypothenuse AC, BO 
the radius of the cir- 
cumscribing circle. 

(2) Divide twice the area of 

the triangle by the 
hypothenuse = BD, or 
(600 X 2) -r- 50=24. 




FIG. 23. 



(3) OD=VBO a — BD*=7; DC = OC— OD, or 18. 



(4) BC = V'BD*+DC a =30. 



(5) AB=VAC a — BC»=40. 



ABC is a 30, 40 and 50 triancle. 



152 



FAIR CHILD'S SOLUTION BOOK. 




FIG. 24. 



Second Solution. 

(1) Let ABC be the right tri- 

angle. 

(2) Construct the square KC 

upon its hypothenuse, and 
from the corners of this 
square let fall the perpen- 
diculars KE, LH and BG, 
this forms three new tri- 
angles, equal to ABC, and 
a square GE. 

(3) Now, through the corners of 

the square KC, draw lines 
parallel to the perpendicu- 
lars already drawn, and we 
have the square DN, each of the triangles similar to 
ACM and equal to ABC. 

(4) Therefore, □I3N=nKC+4A ABC = 4900. 

(5) The DGE = DKC— 4AABC = 100. 

(6) Hence. DF=AB+BC = V4900=70. 

(7) HG = AB— BC=V'i00. 

(8) Hence, AB=£(FD— BE)+EB=40. 

(9) BC=AB— EB = 30. 

.-. ABC is a 30, 40 and 50 t/iangle. 

PROBLEM 326. 

The area of a right triangle is 924 sq. ft., the sum of the base and 
perpendicular is 89 ft.: find the sides of the triangle. 



Solution. 



(1) 

(2) 



From Fig. 24, ABC^the triangle. 
MN=AB+BC=89, MN 2 =7921 sq. ft. 



(3) KLa = DN-4AABC = V4225, or 65 ft.=AC. 

(4) GE = KC— 4AABC = 529 sq. ft., EB= V529 = 23 ft. 

(5) BC=i(FN— GB)=33ft. 

(6) AB=AE, or BC+BE, or 56 ft. 

.-. ABC is a 33, 56 and 65 triangle. 
PROBLEM 327. 



Show that the difference between the hypothenuse and the sum of 
the two legs of a right-angled triangle is equal to the diameter of the 
inscribed circle. 



MENSURA TION. 



153 



Solution. 

(1) Let ABC be the triangle, 

and O the center of the 
inscribed circle. BE, 
BF, AE, AK, KC and 
FC are tangents to the 
same circle; hence EB 
= FB, AE = AK and 
KC=FC, also AE= 
OE=OK. 

(2) EB+KC=FB+CF. 

(3) EB+KC + AE + AK— (FB+CF) = AE+AK. 

.". AF+AK = FO + OE, the diameter of the circle. 
Note. — Solved bv the author for the Teachers' Review. 




FIG. 25. 



PROBLEM 328, 

From a right-angled triangular piece of tin whose sides measure 
a = 13, b = 12 and c = 5 inches, cut out. first, the greatest possible cir- 
cle; and after that, the greatest possible rectangle: find the length and 
breadth of the rectangle. 

Solution. 




FIG. 26. 

(1) Let ABC represent the tin triangle, and OD — AD, the 

radius of the inscribed circle. 

(2) Join OC, cutting the circumference of the circle in T. 

(3) From T draw TG parallel to AC, and TP perpendicular 

to AC. TG and TP will be the length and breadth re- 
spective!}' of the required maximum rectangle. 

(4) The perimeter of the triangle ABC=30 in. 

(5) The radius of the inscribed circle is found by dividing 

twice the area by its perimeter, or (12 X 5) -4-30=2 in. 

(6) DC = AC— AD=1(> in. 



154 FAIRCHILD'S SOLUTION BOOK. 



(7) OC=V(DC 2 +OD 2 ) =10.1981 in. 

(8) TC = OC— OT = 8.1981 in. 

(9) Then by sfmilar triangles, OC : OD :: TC : TP, or 

10.1981 : 2 :: 8.1981 : (TP = 1.6077 in.). 

(10) Also, AB : AC :: GH : HC, or 5 : 12 :: 1.6077 : (HC= 

3.8584 in.) 

(11) OD : DC :: TP : PC, or 2 : 10 :: 1.6077 : (PC=8.0388 in.) 

(12) .-. PH = PC— HC=4.1804in. 

.-. 1.6077 in.r=the breadth, and 4.1804 in.=the length of 
the rectangle. 

PROBLEM 329. 

To determine a right-angled triangle, haviug given the hypothe- 
nuse h = 100, and the difference of the two lines drawn from the two 
acute angles to the center of the inscribed circle = d or 20 V 5. 

Solution. 

(1) Let ABC=the triangle, 

AO and OB the two 
lines drawn from the 
acute angles to the cen- 
ter O, x— A E, EB = 100 
— x, and r^OE, the 
radius of the inscribed 
circle. 

(2) A0 2 =i 2 +r 2 , and OB fig. 27. 

= (100— .r) 2 -f-r 2 . 

(3) /. ;r 2 +r 2 — [(100— *) 2 +r 2 ] = 20V5 squared, or 2000. 

(4) Squaring and reducing, 200^=12000. We find *=60. 

(5) AL = AE, LO=OD = LC, and EB = BD. EB = AB— AE, 

_or 100— AE, or 60=40. 

(6) EB 2 +OE 2 , or 40 2 +r 2 =20V5 = OD 2 . 

(7) r=20. AL+LC=80. BD-fDC = 80. 

/. ABC is a 60, 80 and 100 triangle. 
Note.— d, or 20 V 5"= VaO 2 — OB 3 . 

PROBLEM 330. 

The perimeter of a right-angled triangle is 240 ft., and the radius 
of the inscribed circle is 20 ft.: find the sides of the triangle. 

Solution. 

(1) Fig. 27. The perimeter of the triangle ABC is 240, LC, 
or CD is 20, the radius of the inscribed circle. 




MENSURA TION. 



155 



(2) LC+CD=40, and 240—40=200, the sum of the lines AL, 

AE, EB and BD, which are tangents to the same circle. 

(3) Hence, AL=AE,EB = BD, and AL+DB = AE+EB=100, 

the hypothenuse. 

(4) AAOB=AAOL+ABOD, because they have the same 

bases and altitudes. 

(5) P being the middle point of AB, CP = 50. 

(6) The area of the triangles AOL, BOD and AOB is 

[ (100 X 20) H-2]X 2=2000. 

(7) The area of LU is 20^=400. 

(8) The area of AABC is 2000+400 = 2400, and its altitude 

CH = (2400X2) - rl00=48. 

(9) PH = V( 50 2 — 48 2 )= 14; BH=50— 14 = 36; 

BC = V(48*+36 2 )=80. 
.'. The sides of the triangle are 60, 80 and 100. 
Note. — This solution was prepared by the author for the School 
Visitor. 

PROBLEM 331. 
A tree 120 ft. in height, standing on a level, is broken off so that 
the part broken off is 30 ft. longer than the stump. If the parts remain 
in contact, what is the length of the path through which the top of the 
tree passes in falling to the ground? 

Solution. 

(1) Let AC represent the tree, 120 ft., 

B the point where the trunk 
broke, and AB= (120— 30)-^2= 
45 ft., the stump. 

(2) Then, BC, the part broken off, is 

75 ft. 

(3) The arc CD is the path described 

by the top, C. 

(4) By trigonometry, ZABD = 53°7'50"; 

hence, CBD=126° 52' 10". 

(5) As 360° is to the number of de- 

grees in the arc (126m), so ^ s 
the circumference of the circle 
(150Xtt) to the length of the arc, 
CD, or 166.07 ft. 

NOTE. — This solution was prepared by the author for the School 
Visitor. 

PROBLEM 332. 

A statue c ft. high stands on a column d feet high: how far from 
the foot of the column must a boy stand that the statue may subtend 
the greatest possible visual angle? 




FIG. 28. 



166 



FAIRCHILD'S SOLUTION BOOK. 



(1) 



(2) 



(3) 



(4) 



Solution. 

Let BD^, the height of the col- 
umn; CD=<^, the height of the 
statue; and .#=AB, the distance 
the boy stands from the foot of 
the column. 

Describe a circle which shall pass 
through D ano C, and tangent 
to AB, A being the point of 
tangent/. 

The angle DAC being an inscribed 
angle, is greater than if A were 
moved either way on AB, for it 
would be exterior to the circle. 

Since angle A is measured by half 
angle C by half of the same arc, 
triangles ABD, ABC. 




FIG. 29. 



of the arc AD, and 
we have the similar 



(5) c : x :: x : c-\-d. Whence, x=.yj [c(c-{-d) ]. 

(6) Now, A is the point at which the boy can see the statue 

at the best advantage. Let r=120 ft, and d=&) ft.; 
then #=146.96+ ft. 

PROBLEM 333. 

Find the three sides of a right-angled triangle whose perimeter is 
120 ft., and whose base is § of its perpendicular. 

Solution. 

(1) The base and perpendicular being as 3 to 4, the hypoth- 

enuse is V (3 2 -{-4 2 ) = 5, proportionally. 

(2) This triangle whose perimeter is 3-}-4-r-5 = 12, is similar to 

the triangle whose sides are required; hence, 120-7-12, 
or 10 times these, are 30, 40 and 50 ft., the required 
sides. 



PROBLEM 334. 

Required the sides of a right triangle which shall contain the 
greatest area under the shortest perimeter, when the square of the area 
shall be equal to the product of the three sides. 

Solution. 



(1) Let ;t— the base, y = the perpendicular; then, V (jr 2 -f-jj> 2 ) = 

the hypothenuse. 

(2) By the conditions we must have %x 2 y 2 =zxyyj (* 2 +7 2 ), or 

xy = 4^/(x 2 +j/z). 

(3) Squaring, we have # 2 jj/ 2 =16# 2 -)-16> 2 , which gives x=4y-±- 



MENSURA TION. 15'i 



VCr 2 -i6). 

(4) Substituting this value of x in the value for the area, we 

have 2j 2 -i~V(j /2 — 16), a maximum. 

(5) Differentiating, dividing by dx and equating to zero, we 

have 2y 2 =64, or j/=4V2, and x=±^J2- Whence the 
hypothenuse = 8. 

PROBLEM 335. 
The diameter of the circumscribing- circle of a right triangle added 
tojthe diameter of the inscribed circle, is 71 f, and the perpendicular is 
\ of the base: find the sides. 

Solution. 

(1) The hypothenuse is the diameter of the circumscribing 

circle. Hence, the sides of the triangle are some mul- 
tiple of 3, 4 and 5. 

(2) Let AB=3, BC=4, and AC = 5. 

(3) The diameter of the inscribed circle is 2(ABxBC)-f-(AB 

+BC+AC)=2. 

(4) .'. The diameter of an inscribed circle is equal to 4 times 

the area of the triangle divided by its perimeter. 

(5) The diameter of the circumscribing circle is ABxBCX 

AC = diameterx2 times the triangle; or diameter=(AB 
BCX AC)-i-twice the area of the triangle, or 5. 

(6) Hence, the sum of the diameters=7. 

(7) In the problem it is 1 ^ L . .'. The required is \ l , and the 

sides are 30f, 41 and 51 \. 

NOTE. — We should remember that the continued product of the 
three sides of a triangle divided by twice the area of the triangle is 
equal to the diameter of the circumscribing circle. 

PROBLEM 336. 
Two men, A and B, started from the same point at the same time; 
A traveled southeast for 10 hr., and at the rate of 10 mi. per hr., and 
B traveled due south for the same time, going 6 mi. per hr : they 
turned and traveled directly towards each other at the same rates 
respectively, till they met: how far did each man travel? 

Solution. 

(1) Let C be the starting 

point, CA=the distance 
A travels southeast, and 
CB = the distance B 
traveled south. 

(2) Then CA = 100 mi., and 

CB = 60 mi. 

(3) Now draw AD perpendic- 

ular to CB produced to D. KT(; . 30. 




158 



FAIRCHILD'S SOLUTION BOOK. 



(4) As the Z D=a right angle, and L C=45°, then CD=AD % 

(5) 2AD 2 =AC 2 = 100 2 , whence AD=50V*2, and BD=50V2 

—60. 



(6) In the right triangle ADB , AB = V (50y2) 2 + (50y2— 60)- 

= V 13600 — 6000V 2= 71.517261+ mi. 

(7) A and B together travel 16 miles per hour, and the time 

required, until they meet in traveling AB, is T V of AB = 
■4.469228+ hr. 

(8) .'. A traveled 44.69828+ mi., and B, 26.81897+ mi. of the 

distance AB. 

.*. The total distance traveled by A is 144.69828+ mi., 
and by B, 86.81897+ mi. 

Note. — This solution was prepared for the American Mathematical 
Monthly by G. B. M. Zerr, A. M., Ph. D., President of Russell College, 
Lebanon, Va. 

PROBLEM 337. 

On a globe 20 ft. in diameter, 30 ft. above a plane, there is a bo\ r 4 ft. 
tall to his eyes: how much surface on the plane is hidden from him? 



Solution. 



(1) 



30 ft. = EB, the distance the globe 
is above the plane; 20 ft.— DE, 
the diameter of the globe; and 
AD = 4 ft., the distance the boy's 
eye is above the ball. 

(2) 14 ft.=AD+DL = AL. 

(3) AF=VAL 2 — LF 2 =4V6~ft. 

(4) Now, by the similar triangles ALF 

and ABC, AF : LF::AB : BC, 
or 4V6 : 10 :: 54 ^BC, from 
which BC = 540+4 V6=135+V 6 

=135V^6xi-= 1 -^^ ft. 



6 




(5) 



■BC 2 = 



A35V6 



= area of the 



FIG. 31. 



surface hidden from the boy. 



PROBLEM 338. 

What is the area of the largest square that can be inscribed in a 
semicircle, the diameter of the circle being- 20 inches? 



MENSURA TION 



159 



Solution. 

(1) Let AB be the diameter of 

the circle. 

(2) Draw SB equal and perpen- 

dicular to AB, tangent to 
the circle; E and SO to the 
center of the circle. 

(3) Let K be the point where SO 

cuts the circle. 

(4) From the point K, draw KC 

parallel to SB; TK parallel 
to AB, TX parallel to KC. 
Hence the square is com- 
pleted. 

(5) Then 1 ^K^40C 2 -hOC 2 , or 50C 2 . 

(6) ButOK 2 =100; .\ 5OC 2 = 100; OC 2 

or 4.472+ in. 

(7) CX = the side of the square, or 4.472 X 2=8 944+ in 

(8) CX 2 =79.99+ sq. in., area of the inscribed .square. 




:20. 



d OC = V20, 



PROBLEM 339. 

I have an inch board 5 ft. long - , 17 in. wide at one end and 7 in. at 
the other: how far from the larger end must it be cut straight across 
so that the two parts shall be equal? (A\ H. A., p. /<>-, prod, /of.) 

Solution. 

(1) Let ABCD be the board. 

(2) DC=7 in., the width of the small end, 

and AB = 17 in., the width of the 
large end. 

(3) CB = 5 ft., or 60 in., the length of the 

board. AL = 10 in. 

(4) Then, by similar triangles, AL : AB : : 

DL : KB, or 10 : 17 :: 60 : (BK= 
102 in.). 
The area of ZDCK=(7x 42)+2 = 147 sq. 




FIG 



(5) 

(6) The area ofZABK=(102xl7) 



867 sq. in. 
=720 sq. in. 



(7) Then, the area of DABC=867— 147 

(8) Area of FECD = 720+-2 = 360 sq. in. 

(9) The area of FEK = 360+ 147=507 sq. in. 

(10) .-. 867 : 507 : : 102 2 : KE 2 . EK=78 in. 

(11) 102 in.— 78 in. =24 in. 

.*. EB = 2 ft., the distance the board must be cut from the 
large end. 



160 



FAIRCHILD'S SOLUTION BOOK. 



PROBLEM 340. 

The sides of a triangular field are 39, 40 and 50 rd. : from what 
point in one of the shorter sides must a line be drawn parallel with the 
other shorter side, so as to divide the field into two equal parts? 



(i) 



(2) 

(3) 

(4) 
(5) 



Solution 

Let ABC be the triangle, 
and draw EK parallel to 
AB, cutting off the half 
of AB or [(40X30)-r-2] 
^-2 = 300 sq. rd. 

Then we have the propor- 
tion, 600 : 300 : : 40 2 : 
(EK, or 28.284 rd.) 




FIG. 34. 



EK=28.284 rd., length of the line that divides the tri- 
angle into equal parts; and it is parallel to AB, and AB 
=40 rd. 

Also, draw TP parallel to CB, whose side is 30 rd. 

.\ 600 : 300 :: 30 2 : CB 2 , from which CB=21.213 rd. 

.-. TP=21.213 rd., dividing the field into equal parts par- 
allel to CB. 



PROBLEM 341. 

Two trees stand on opposite sides of a stream 40 ft. wide; the 
height of one tree is to the width of the stream as 8 is to 4, and the 
width of the stream is to the height of the other as 4 is to 5: what is 
the distance between their tops? 

Solution. 

(1) From Fig. 9, AB=40 ft., the width of the stream. CB 

and FA = the height of the trees. 

(2) Then by the conditions of the problem, one tree is twice 

the width of the river. 

(3) .-. CB=80ft. f FE=40ft„ FA=50 ft., the height of the 

2d tree. 

(4) CE = 30 ft.; FC, the distance between their tops= 

V^FE 2 +CE 2 )=50ft. 
.'. The required distance is 50 ft. 



PROBLEM 342. 

The lengths of the lines that bisect the acute angles of a rig4tt- 
angled triangle being 48 and 60 respectively, to determine the three 
sides of the triangle. 



MENSURA riON. 



161 



Solution. 

(1) Let BC=*. AC=nx, AB= 

mx, AD=48=a, BE=60= 

b. 

(2) AB 2 =AC2 + BC 2 , or m*x* 

—n 2 x 2 -\-x' 1 , or w 2 =tt 2 +l. 

(3) .-. m 2 —n 2 =l ... (1). 

(4) Area ABC = area BAD + 

area DAC. 

(5) Area ABC^ABX ACsinA 

== ^mnx^sinA = mnx 2 sin^A- 
cos^A. 

(6) Area BAD=|ABx ADsinDAB^^w^sin^A 

(7) /. DAC=|ACxADsinDAC=^^sin|A. 

(8) .\ mnx^sin^Acos^A^^amxs'inA^+^atixsin^A. 

(9) .*. 2mnxcos\A=am-\-an. 

AC nx 




(10) From triangle, DAC = cos|A = 



AD 



(11) .*. am-\-an 



2mh 1 x 



■ (2) 



(12) Similarly from area ABC = area ABE+area CBE. 

(13) We find *+«*==^p . . . (3). 

(14) <8)-H3) gives g+^W... (4). 

(15) Eliminating n, using (1) and (4), and calling -j^— d =.64. 

(16) w 6 +2/« 5 — {l+2d)m*— (4+2</)w 3 — (\-2d)m 1 + 2(1 + 

(17) Restoring numbers, w 6 +2w 5 - 2.28;« 4 — 5.28w* 3 +28/« 2 + 

3.28^+1.4096=0. 

(18) Reducing by Horner's method to 2 decimal places, m = 

1.25; then a=.75. 

(19) Substituting these values of m and n in (2), ,r=57.24+ 

^=42.93+, and w*=71.55+. 

.-. The sides of the triangle are, AC = 42.92+, BC = 
57.24+, and AB=71.55 + . 



PROBLEM 343. 

A barn is 60 ft. long, 20 ft. high to the square; the rafters, exclu- 
sive of eave projection, are 25 ft. long, and the roof is f pitch: find cost 
of siding at $20 per M. 



162 FAIRCHILD'S SOLUTION BOOK. 

Solution. 

(1) From Fig. 36, let ABC be the gable end of the barn, 

and AC = BC=25 ft., the length of the rafters. 

(2) By the conditions of the problem,CD = 3 ft., and AB, the 

spanz=8 ft. 

(3) Then, AD=4 ft., that is, we suppose that ABC and ADC 

are similar to ABC, the given gable end. 

<4) AC=V(AD2+Cl)2)=5, whence ADC is a 3, 4 and 5 
right triangle, or the smallest integral triangle. 

(5) Now, as AC is 5 in the small triangle, then we have AB, 
the hypothenuse of the large triangle=5X5 = 25. 

<6) Likewise, AD=20 ft., and CD = 15 ft.; then AB=40 ft., 
the width of the span. 

(7) The area of ABC=£(ABXCD):=300 sq. ft., or 300x2= 

600 sq. ft., the area of both gable ends. 

(8) The barn is 20 ft. high to the square, then the area of 

the square is 40X20, or 800 sq. ft., and 800x2 = 1600 
sq ft., area of both ends, excluding the gable ends. 

(9) 60X20X2=2400 sq. ft., area of both sides of the barn. 
(10) .*. 2400 sq. ft.+1600 sq. ft.+600 sq. ft.=4600 sq. ft., total 

number of square feet in the barn, or 4f M. 
<11) $20X4f =$92, the cost required. 

Note. — A Gable is the triangular portion of the end of a building, 
bounded by the sides of the roof and a line joining- the eaves. 

A Gable End is the triangular-topped end of a house or barn. 

The Pitch is the inclination of a roof. The common pitch has a 
rafter three-quarters the length the span. In the above problem the 
pitch is f of the span AB, or CD=f of AB, or 15 ft. The Gothic has a 
full pitch, the rafters being the length of the span; the Greek has a 
pitch \ to \ of the span; the Roman has a pitch from \ to f of the span; 
and the Elizabethan has rafters longer than the span. 

II. THE EQUILATERAL TRIANGLE. 

PROBLEM 344. 
If the side of an equilateral triangle is 2s, find its perpendicular. 

Solution. 

(1) In the triangle ABC, draw the perpen- 

dicular CD. 

(2) AD^, ADC is a right triangle. 

(3) Then, AC 2 -AD 2 = CDl 

(4) Now, as AC=2.?, we have 4j 2 — * 2 = CD 2 . 

nr\ /IT FIG - 36 - 




MENSURATION. 



163 



Rule. — Multiply half of the side by the V3. 
Note.— It would be well to remember that the V 3^=1.73205. 

PROBLEM 345. 
To find the area of an equilateral triangle having- given the side. 

Solution. 

(1) From Fig. 36 we know that AD is s, or half the side of 

the triangle, and CD=Sy'3. 

(2) Then the area is ADxCD, or sXs*JS=s°~ V3. 

Rule. — Multiply the square of half the side by V3. 

PROBLEM 346. 

In an equilateral triangle having given the lengths of the three 
perpendiculars, drawn from a certain point within, on the three sides; 
to determine the perpendicular. 

Solution. 

(1) Let ABC represent the equi- 

lateral triangle, and DE, 
DF, DG, the given per- 
pendiculars. 

(2) From the points D, draw 

the lines DA, DB, DC to 
the angular points; and let 
fall the perpendicular CH 
on the base AB. 
(8) Put the three given perpen- 
diculars, DE=a, DF=£, 
DG=f, and put i=AH, or 
BH, half the side of the fig. 37. 

equilateral triangle. 

(4) Then AC or BC=2j, and by right angles, the perpen- 

dicular CH = VAC 2 — AH 3 = V4i 2 — 5 2 = V3^=^V3T 

(5) The area of a triangle is expressed by \ the product of 

the base and height, and a triangle is equal to half a 
rectangle of equal base and height. 

(6) .*. The whole triangle ABC is=^ABxCH=^XiV3= 

* 2 V3. 

(7) The triangle ADB=|ABx DG=sXc=cs. 

(8) The triangle BCD = ^BCX DE=sXa=sa. 

(9) The triangle ADC=%ACxFD=sXb=bs. 

(10) But the three last triangles are equal to ABC, that is 5 a V3 
— as-\-bs-\-cs y hence dividing by s, s-J3 = a-\-b-\-e. 




164 



FAIRCHIL&S SOLUTION BOOK. 



:. 5V3 = HC=FD+DG+ED. 



PROBLEM 347. 

Three horses are tethered each to a rope 15 rd. in length to the cor- 
ners of an eqilateral triangle whose side is 30 rd: over how much space 
can they graze, and how much remains ungrazed? 

Solution. 

(1) Let ABC be the triangle. CFD, 

BDE and AFE=the three 
sectors grazed over by the 
horses. 0=space ungrazed. 

(2) Since the angle of an equilateral 

triangle is 60°, because the 
three angles of every triangle 
are equal to two right angles, 
each sector is \ of a circle. 
The three sectors=^ of a 
circle. 

(3) Area of circle, radius 15 rd. = 15 2 X7r = 2257r sq. 

(4) 225k-^2^ 353.43 sq. rd.^area grazed over. 

(5) Area of CBA=BD a V3, or 15 2 V 3=389.711 sq. rd. 

(6) 389.711 sq. rd— 353.43 sq. rd. =36.281 sq. rd., part not 

grazed over. 




FIG. 38. 



rd. 



PROBLEM 348. 

A goat is tethered by a cord 30 ft. long, to the corner of a shed in 
tthe form of an equilateral triangle, side 20 ft.: find area over which it 
may graze outdide of the shed. 

Solution. 

(1) Let ABC be the shed. 

(2) The goat can graze with 

the 30 ft. cord over a cir- 
cle, except a sector of 
60°, or angle A. 

(3) EMD=300°, or JU=f oi a 

circle, which has 30 2 ?rX| 
=2356.2 sq. ft. 

(4) Angle DCS=120°, and 

angle EBS = 120°. 

(5) Angle EBS+angle DCS= 

240°, or | of circle with a 
radius of 10 ft. 

(6) .'. The area of BES+SCD=10*irX| -=209.44 sq. ft. 




FIG. 39. 



MENSURATION. 



165 



(7) Whole area grazed over is 2356.2 sq. ft.+ 209.44 sq. ft.= 
2565.64 sq. ft. 

PROBLEM 349. 
At 80 cents a rod, what would be the cost of fencing- a field in the 
form of an equilateral triangle, its altitude being 6 rd.? 

Solution. 

(1) Let ABC be the triangle. 

(2) AO bisects the angle A, then 

angle OAD=30°. Then AOD 
is a right-angled triangle. 

(3) Now, OD+DE = CO, or the ra- 

dius of the circle. 

(4) CO or AO=| of 6 rd. = 4 rd. 

(5) AB = AOV3, or 4V3=6.92820 

rd. 

(6) The perimeter is 6.92820x3= 

20.78460 rd. 

(7) 20.78460 X. $80 = $166.2768, cost 

of fencing the triangle. 




PIG. 40. 



NoTK.— As AOD is a right triangle and the angle OAD=30°, then 
AO=4 and OD=2 rd. It follows that the hjpothenuse is twice the side 
opposite the acute angle of a right triangle which is 30°. Also if the 
hjpothenuse be given, then the side opposite the 30° angle is half of 
the hypothenuse. 

Prove that AOV3=AB. 

A figure with six sides is called a 
hexagon; the side is equal to the radius 
of the circumscribing circle. Now, if 
I unite the alternate angles of the reg- 
ular hexagon, as AB, BC, and CA, I 
have a regular triangle, called an 
equilateral triangle. Join AD, then 
ADB is a right triangle. Then OD= 
DB._ 

_ AB 8 =AD a — DB 8 =4AO a — AO», or 
(DB 8 )=3AO*. Whence AB=AOV3; 
that is, the side of an equilateral tri- 
angle is equal to the radius of the cir- 
cumscribing circle multipied by the PIG. 41. 
square root of 3. 

PROBLEM 350. 

The area of an equilateral whose base falls on the diameter and 
whose vertex falls on the middle point of the arc of a semicircle, is 80 
sq. ft.: what is the diameter of the semicircle? 




166 



FAIRCHILD'S SOLUTION BOOK. 




FIG. 42. 



Solution. 

(1) Let SPC be the semicircle, 

and ABC the triangle. 

(2) CD is the radius of the 

circle, also the perpen- 
dicular of the triangle. 

(3) The area of an equilateral 

triangle is found by 
squaring half the side 
and multiplying by *J3, 
then we must get the 
side by reversing this 
operation. 

(4) V (81H-1.732) X 2=13 59 ft- 

(5) Since the perpendicular equals half the side multiplied 

by V^< we have 13.59-I-2X V^ — H-77 ft., perpendicular, 
or radius of the circle. 

(6) .-. The diameter is 11.77X2 = 23.54 ft. 

PROBLEM 351. 
John has a garden in the form of an equilateral triangle; from the 
corners to a spring- within the garden, the distances are a=20, 6=28, 
c—31 rods: find side of garden. 

Solution. 

(1) Let ABC be the given tri- 

angle, O the spring. 

(2) PutAO = a=20. BO=£=28. 

CO=*=31. 

(3) Let^=side of triangle ABC. 

The area=|5 2 V3. 

(4) Draw AFB^AOB. With 

B as a center and b as a 
radius, describe the circle 
FOD. With C as a center 
and fas a radius, describe 
the circle DOE. Join EF, 
FD and DE. fig. 43. 

(5) Now angle EAF=2 times angle CAB = 120°. 

(6) /. EFis equal to a side of an equilateral triangle inscribed 

in a circle whose radius is a. .'. EF=^V^- 

(7) In the same manner find FD=^V3, and ED=^V3- 

(8) Area of triangle EAF = 100V3, of triangle FBD = 

196V3. of triangle DCE=240JV^. and of triangle EFD 
=823.18129 sq. rd. 




MENSURATION. 



16? 



(9) .-. The area of the entire polygon AFBDCE=1731.9935 
sq. rd. 

(10) But this is double the area of triangle ABC. .'. Area of 

triangle ABC=875 996785 sq. rd. 

(11) Equating the two expressions for the area, we have 

tj»V3=875.996785 sq. rd. 

.-. j=44 97874 rd., the side of the triangle ABC. 
Note. — Solved by the author for the Teachers' 1 Review. 



PROBLEM 352. 

A deer is tethered from one corner of an equilateral wall whose side 
is 100 ft., by a rope 175 ft. long - : over what area can it graze? 

Solution. 



(i) 

(2) 



(8) 




(6) 



(7) 



FIG. 44. 



Let AOB be the triangle. 
We see that it can graze 

over |- of (175 2 X*-)=: 

80176.06 sq. ft. 
Next over two sectors 

A TF and BET. 

(4) We also have a right tri- 

angle BDT having BD = 
50 ft., and TB 75 ft., to 
find the angle DBT. 

(5) By triecmome'r/, ^DBT 

=48 s VoV. and ZEBT 
=711841°. and ZIBE+ 

ZrAF=i43}ia°. 

From this we have 360° : 

148ltt)°::75*X* : (7049.96 sq. ft.), the area of the two 

sectors. 
It also grazes over the triangle ABT, whose base is AB^= 

100 ft., and altitudeTD or V('5 2 — 50 2 )=25V^ft. 

(8) This gives (100x25V5)^2=2795 08 sq. ft. 

(9) Total area grazed over is 80176.06+7049.96+2795.08= 

90021.10 sq. ft. 

PROBLEM 353. 

I have a lot in the form of an equilateral triangle whose sides are 
100 ft. At each corner stands an orange tree; the height of the first 
tree is 10 ft., the second 20 ft., and third 30 ft. At what distance from 
the foot of each tree must a ladder be placed, so that without moving 
it at the base it may reach the top of each? Find the length of the lad- 
der. 



168 



FAIR CHILD'S SOLUTION BOOK. 




FIG, 45. 



Solution. 

(1) Let ABC be the garden, 

and AX, BP and CE the 
trees at the corners. 

(2) Join the tops of the trees 

by the lines XP, PE and 
EX. From W, T and V, 
the middle points of XP, 
PR and EX, draw WL, 
1 K and VN perpendic- 
ular to XP, EPand EX, 
and at L, F and N draw 
perpendiculars to AB, 
BC and CA in the equi- 
lateral triangle ABC, 
meeting at O. 

(3) Then O is equidistant from X, P and E; L is equidistant 

from X and P; and OL being perpendicular to the 
plane ABPX, every point of OL is equidistant from X f P. 

(4) For the same reason OF and ON are equidistant from 

E and P, F and X. 

(5) Therefore, O, their point of intersection, is equally dis- 

tant from X, P and E. 

(6) Draw XK, PZ and XY perpendicular to PB, EC, and 

also is XY perpendicular to EC. 

(7) Now, if the student refers to Fig. 

7, it will be easy to understand 
Fig. 45. Let us examine Fig. 46 

(8) Let A'B'C be the triangle whose 

sides are 100 ft., L=point L', 
S=S', F=F', etc. 

(9) Draw F'D parallel to O'L', or to 

C'S', and R'O parallel to DL; 
then the two triangles F'DB'and 
O'R'F' are similar. 

(10) We know that S, R and M are the middle points of AB, 

BC and CA, and that the triangles XBP and WSL, PEZ 
and VMN, XYE, VMN are similar. 

(11) The figure AXPB is a trapezoid, and WS=|(AX+PB) 

= 15 ft., TR=|(EC+PB)=:25 ft., and VM=i(EC+XA) 
=20 ft. 

(12) By similar triangles, XK : PK:: WS : SL, from which 

SL=1| ft. 

(13) PZ : EZ::TR : FR, or FR=2| ft. 

(14) YX : EY::VM : NM, or MN=4 ft. 




FIG. 46. 



MEN SURA TION. 



im 



(15) From this we find LB = 48£ ft, AL=51^ ft., CF = 47^ ft., 

FB:=52£ ft., CN=46 ft., and NA=54 ft. 

(16) ZS'C'B'=30°, also /DF'B', as F'B'=FB, or 52| ft. 

(17) From t he note under Fig. 40, DB'=| of F'B, or 26£ ft. 

(18) FD=V(FB'2-fBT) 2 )=54.466ft. 

(19) R'O', or DL'=L'B'— B'D=22i ft. 

(20) By similar triangles, F'D : DB'::R'0' : F'R', or F'R'= 

12.99 ft. 

(21) R'D, or O'L'^DF— F'R=32.47+ ft. 

(22) B'0'=^(M>+DB*) =58.36+ ft. 

A'O f =V(STL' a +T7A' 2 ) =60.88+ ft. 

(23) By similar triangles, DB' : F'B':: F'R': F'O', or F'0'= 

25.98 ft. 

(24) CO , =V(C 7 F' a +O r F'2)=:54.15 ft. 

OE = ^"(CkT'+CE 2 ) =61 7+ ftf length of the ladder. 

.-. AO=60.88-h ft., OB=58.36-f ft., CO=54.15+ ft., and 
61.7+ ft. = the length of the ladder. 



PROBLEM 354. 

On the three sides of any plane triangle construct equilateral*; 
join the centers of these triangles; prove that the resulting triangle Is 
equilateral. 

Solution. 

(1) Let ABC be any tri- 

angle, having the 
equilateral triangles 
ABK.BDCandACE 
described on AB, 
BC and CA. 

(2) Let G, H, F be the 

centers of these tri- 
angles. 

(3) Now, we are to prove 

that GHF is an equi- 
lateral. Circum- 
scribe the three tri- 
angles and GHF by 
circles. 

(4) Draw SPH perpen- 

dicular, FAC rather 
at right angles with 




FIG. 47. 



AC, to intersect AF if produced, in S. 



170 FAIRCHIL&S SOLUTION BOOK. 

(5) AS passes through F; then it bisects /FAC, which makes 

ZXAC = 30°. 

(6) In like manner angle QCA=30°, from this it is easy to 

see that SFP is an equilateral, and angle FPH=120°. 

(7) .\ FGHP is a quadrilateral and the sum of their opposite 

angles = 180°, as angle FPH=120°. Angle FGH = 180° 
— 120° = 60°. In the same manner it may be proven that 
HFG and FHG are each equal to 60°; therefore FGH 
is equilateral. 

Note. — This solution was prepared by the author f or the American 
Mathematical Monthly. 

PROBLEM 355. 

A wheel 8 ft. in diameter is run by means of a belt from a wheel 1 
ft. in radius; the distance between their centers is 12 ft., and the belt 
is crossed between them, causing- them to revolve in opposite direc- 
tions: what is the length of the belt used? 

PROBLEM 356. 

A ladder 20 ft. long - leans against a perpendicular wall at an angle 
of 30°: how far is its middle point from the bottom of the wall? 

Solution. 

(1) If the ladder makes an angle of 30° with the wall, it 

must make an angle of 60° with the base. 

(2) Suppose on the opposite side of the wall a similar lad- 

der be placed, and meet the wall at an angle of 30°, the 
angle at the base wouid be 60°. 

(3) Then, with the two ladders and the base line, we have 

an equilateral triangle, each side 20 ft. and each angle 
60°. 

(4) Now, joining the middle points in the sides of any equi- 

lateral triangle divides the triangle into four similar 
triangles. 

(5) The middle points of the base line and of the ladder are 

the middle points of the equilateral triangle. 

(6) Hence, the line that joins these points is 2CK-2, or 10 ft. 

PROBLEM 357. 

An equilateral triangle and a square have equal areas: find the 
ratios of their perimeters. 

Solution. 
(1) As the area of an equilateral triangle equals \ of the 
square of a side multiplied by y/3, or 1.73205, we find 
the side by reversing the operations in this rule. 



MENSURATION. 171 



(2) When the area is 1, the side is ^(l-r-1.73205) X4=£, 

nearly, and the perimeter is 3xf=2^. 

(3) The side of the square whose area is 1, and the perimeter 

4. 

(4) The required ratio, then, is 2\ to 4, nearly, or 9 to 16. 

PROBLEM 358. 
The area of one equilateral triangle is ^ of another; if the radius 
of the inscribed circle of the larg-er triang-le is 8 inches, what will be 
the circumference of the circle inscribed in the smaller? 

Solution. 

(1) Ratios of any homologous lines, such as sides, altitudes, 

radii of inscribed or circumscribed circles, is Vi 9 6~ = i- 

(2) Then the radius of the required circle is £ of 8, or 6 in., 

and its circumference is 6x27r or 12ir. 

PROBLEM 359. 
A park of three equal sides has a drive along" the boundary occupy- 
ing .19 of the whole park; if the nearest distance from the drive to the 
center is 9 rd., what is the area of the inscribed circle? 

Solution. 

(1) The area of the park inside of the drive is .81 of the whole 

park. 

(2) The 9 rd. must be V-81, or .9 of the nearest distance from 

the center of park to its side. 

(3) 9 rd.-f-.9 = 10 rd., said distance, which is the radius of the 

inscribed circle. 

(4) Area of circle is 10 2 7r, or 1007r. 

FORMULAS USED IN SOLUTIONS. 

1. Area = one-half the side squared and multiplied by yJS — 
1.73205. 

2. Side = V (area^V^) X2. 

3. Altitude = one-half the side multiplied by V^- 

4. Radius of inscribed c;rcle=one-third of the altitude. 

5. Radius of circumscribed circle = two-thirds of the alti- 
tude. 

6. All equilateral triangles are similar. 

7. Each angle is J of a circle of 60°. 

8. Side=the radius of the circumscribed circle multiplied 

by V3. 

Note — The student should commit the above rules; it will often 
be the saving of much labor. 



172 FAIRCHIL&S SOLUTION BOOK. 



III. SCALENE TRIANGLES. 

PROBLEM 360. 

What is the area of a triangle whose base is 20 ft., and altitude 12 
ft.? 

Solution. 

(1) 20 ft.=base. 

(2) 12 ft. = altitude. 

(3) (20X12)^=120 sq.ft. 

.*. 120 sq. ft. = area of the tri- 
angle ABC. 

FIG. 48 

Bulo. — Multiply the base by the altitude and take half the product. 

PROBLEM 361. 

Find the area of a triangle whose sides are 13, 14 and 15 ft. respec- 
tiyely. 

Solution. 

(1) 13 ft.+14 ft.H-15 ft.=42 ft., sum of the 

sides. 

(2) $ of 42=21 ft., half the sum of the sides. 

(3) 21 ft.— 13 ft.=8 ft., 21 ft.— 14 ft.=7 ft., 

21 ft.— 15 ft.=6 ft. 

(4) 21x8x7x6=7056, product of the half 

sum and the three remainders. 




(5) V 7056=84 sq. ft., area of ABC. 




FIG. 4<>. 



Bui©. — Add the three sides together and take half the sum ; from 
the half sunt subtract each side separately; multiply the half sum 
and the three remainders together and extract the square root of the 
Product. 

NoTK. — The demonstration of this rule can be found in any g-ood 
geometry. 

Suppose we were to find the greatest inscribed square in 
the above problem. 

(1) ABC represents the given triangle, GDEF the inscribed 
square. 



MENSURA TWN. 173 

(2) AB=£, CH=^, the side of the inscribed square byx\ then 

will CK—h—x. 

(3) But there may be three different squares inscribed in any 

scalene triangle not right-angled. Let AB = 15, CB= 
14 and AC = 13, or c, b y a. Also, let \{a-\-b-\-c)—s. 



2 



(4) By formula, CH, or k=- i /s{s— c) (s— b(s— a) =ll£. 

(5) Then the perpendiculars upon CB = 12, and AC = 12^J. 

(6) By similar triangles, AB : GF :: CH : CK; that is, c : x :: 

h : k—x. 

(7) ^=^7-7. or 6^* r in. described upon AB; also 6 T % in. on 

CB, and 6jff in. on AC. 

(8) .*. 6^"*^ in., 6 T 6 f in., and 6jf| in. are the sides of the in- 

scribed squares, and it is proven to be true that in any 
acute-angled triangle the inscribed square described 
on the least side of the triangle will be the greatest. 

(9) Hence, 6£J$ m - ' s tne s ^ e °f tne square described on the 

least side of the triangle, which will be the greatest. 

Notb. — This problem was prepared by the author for the Teachers' 
Review. 

PROBLEM 362. 
The area of a triangle is 84 sq. ft.; two of the sides are 13 and 15: 
find the other side. 

Solution. 

(1) From Fig. 48, AB = 15 ft., CB = 13 ft. 

(2) CD=84X2-M5=11.2 ft. 

(3) CDB is a right triangle, with its hypothenuse 13, and 

altit ude 11.2 ft. 

(4) BD = y/(13) 2 — (11.2)2=6.6 ft. 

(5) AD- 15— 6.6=8.4 ft., the base of the right triangle ADC. 

(6) CA = V(H.2) 2 — (8.4)2=14 ft., the third side. 

PROBLEM 363. 
In a triangle whose sides are 13, 14, 15, required the length of each 
line from the angles to a common point, which shall divide the tri- 
angle into three equal parts. 

Solution . 

(1) Let ABC be the triangle. 

(2) We know that the area of the triangle is 84; then each 

part is 28. 

(3) The common point is at the intersection of lines drawn 

parallel to the sides and one-third the distance from 
the side opposite angle. 



174 



FAIRCHILD'S SOLUTION BOOK. 



(4) AC=13, CB = 15, and 

AB = 14. 

(5) EO = (28x2)+14=4. 

(6) FO=(28X2)+13=ff. 

(7) OD=(28x2)-M5=-ff 

(8) AH=Y- A X, or OH 

=Y; ■■■HE= 

l/(OH2— UE a )=f 
A E=y+j=Y; A0= 

T/(AE2_^.QE2)- 

£1/505=7.49073+. 



(9) 




(10) BC=V(BE» + UE») = 

4^/673=8.6474+. 

(11) OG=£ofl5=5. FG=v/OG 2 -TyF2 

(12) CF=«+V=W- 

(13) CO =T /[ (W) " + (») a ]=ii/592=8.11035+. 



FIG. 50. 



8 8 



PROBLEM 364. 

In a triangular garden whose sides are a=300 ft., 6=250 ft., and 
£=200 ft., is a spring equally distant from each corner: how far is it 
from each corner? 

Solution. 

(1) Let BAC be the garden. 

(2) CA=300 ft., BC=250 ft.. 

AB=200 ft., and O rep 
resents the position ol 
the spring. 

(3) BO, OA and OC are radii 

of the circumscribing 
circle and the required 
distance. 

(4) Let g±|±£ =s; then, by 



2 
formula. 



CD= 



-Vs{s- 



a)\s — b) (s — c) — 




248+. 
(5) Now as the triangles CBD, 
CB :CE::CD: CA. 



FIG. 51. 



CAE ate similar, we have 



MENSURATION. 175 

(6) Substituting the values of CB, CD, CA in the above pro- 
portion, we find CE, the diameter of the circle = 302-|- ft. 

.-. BO, AO and CO = 151+. 

NoTB. — This problem was taken from the Teachers' 1 Review. The 
solution there is different from the one above. 

PROBLEM 365. 

A mathematical teacher amuses six of his best students by laying- 
off a triang-ular space on the level school yard, measuring - exactly 
231 ft., 250 ft. and 289 ft. He takes his position in the center, three 
boys occupy positions on the vertices and three girls on the sides equi- 
distant from the teacher. The girls figure the distance they are from 
the teacher, and the boys find out how far each is from the teacher and 
the two nearest girls. {Hancock Co. test ) 

Solution. 

(1) Let ABC be the tri- 

angle, and O the posi- 
tion of the teacher, 
A, B, C the positions 
of the boys, and S, D, 
K that of the girls. 

(2) Let AB=c, CB=a and 

CA=£, and let S= 

i(«+*-N), 

(3) Th e area of ABC — 

l/s(s— a) (S—3) (S—C) FIG. 52. 

= 27720 sq. ft. The 

perimeter of the triangle:=tf-f-£-f-<::=770 ft. 

(4) Dividing twice the area of the triangle by the perimeter 

gives the radius of the inscribed circle, or the distance 
the girls are from the teacher. 

(5) .'. (2770x2)^-770 = 72 ft., the radius of the inscribed 

circle. 

(6) CA = 231 ft., AB = 250 ft. and CB = 289 ft. 

(7) Let *=CS and CK. Then AC— *=SA and CB— *=KB. 

(8) Now, AS, or AC— .*=AD, and KC, or CB— *=DB. 

(9) We have 231 ft.— *+289 ft.— *=250 ft., or ;r=135 ft. 

(10) AS = 231 ft.— 135 ft. =96 ft., and 289 ft.— 135 ft. =154 ft. 

(11) CO =y7 2M I 135^ = 153 ft. AO = V72»+96 a = 120 ft., and 

OB = v^72^+170 2 =170ft. 

Note. — This problem was first published by J. S. Royer, the pub- 
lisher of the School Visitor, and was solved by the author and others. 




176 



FAIRCHIL&S SOLUTION BOOK. 



SCALENE TRIANGLES WHOSE AREAS ARE INTEGRAL. 



4 13 15 

13 14 15 

7 15 20 

11 13 20 
10 17 21 

12 17 25 

13 20 21 
17 25 26 
17 25 28 
13 37 40 



13 40 45 

15 34 35 

15 37 44 

17 39 44 

25 29 36 

25 39 40 

29 35 48 

39 41 50 

13 68 75 

15 41 52 

17 55 60 



20 37 51 

25 39 56 

25 52 63 

25 51 52 

25 74 77 

26 51 55 
29 52 69 

34 65 93 

35 53 66 

36 61 65 

37 91 66 



50 



39 41 

39 85 92 

40 51 77 

41 51 58 
41 84 85 
48 85 91 

50 69 73 

51 52 53 

52 73 75 
43 61 68 



IV. CIRCLE. 

PROBLEM 366. 
Prove that AB is 3 times ET, or KT is \ of AO. 

Proof. 

(1) Let KT=r, SO=R, and ET 

(2) SOK is a right-angled tri- 

angle. 

(3) Then SK=R+r, hypothe- 

nuse. 

(4) OK=2R— r % the perpendic- 

ular. 

(5) SO = R, the base of the right 

triangle. 

(6) R> + (2R— r)*=(R+r)\ 

(7) r=|R, and R=|r. PlG . 53 . 

(8) Now as x=2r, then *=f R. 

Suspose SO, or R=3, then x—\ of 3, or 4. 

(9) Then as SO can be applied on AB 4 times, AB = 12. 

(10) Then ET: AB::1:3. 

(11) Also, KT=2, and AO=6; then KT : AO::l : 3. 

PROBLEM 367. 

From a circular piece of pasteboard whose diameter measures 60 in., 
John cut the larg-est circles that could be measured from it; Mary cut 
the two larg-est circles from what remained: what is the area of one of 
John's and Mary's pieces? 

Solution. 
(1) From Fig. 53, let AB be the diameter of the given circle, 




MENSURA TION. 177 

AO the diameter of one of John's and KT the diameter 
of one of Mary's pieces. 

(2) Then AO=30 in., or S0 = 15 in. 

(3) The area of AO = 15 2 7r=2257r sq. in. 

(4) E0=| of 60, or 20 in. Then, one of Marv's pieces^ 

10 2 tt, or IOOtt. 

PROBLEM 368. 

A circular field is inclosed by 12-foot boards: how many acres in 
the field, if there are as many acres in the field as there are boards in 
the fence surrounding- it, the fence being" 4 boards high? 

Solution. 

(1) Let AB be the diameter of the circle. 

(2) AO=R, the radius. 

(3) 2R.7r=circumference. 

(4) =the number of boards. 

La 

(5) One acre contains 43560 sq. ft. 

R 2 7T 

(6) •'• <or^ — the number of acres in the fig. 54. 

4dobO 

field. 

(7) Since the number of boards in the circumference equals 

t | U C • 4.U C u 8R7T R 2 7T 

the number or acres in the held, 




12 43560" 

(8) Clearing of fractions, 12R=348480, and R=29040. 

(9) -jy- = 193607r, by substituting the value of R. 
/. 193607r = no. of acres. 



PROBLEM 369. 

A gardener is desirous of laying- out a flower garden containing 
10 sq. rd. in the form of a crescent, bounded by a quadrant and semi- 
circle: what is the diameter of the semi-circle? 

Solution. 

(1) The center of the large circle, O, will be on the circum- 

ference of the smaller circle, ADC. 

(2) ACDF is a quadrant. The right triangle ADC=10 sq. rd., 

or AD=the side of a square containing 20 sq. rd. 



178 



FAIRCHILD'S SOLUTION BOOK. 



(3) TheluneorcrescentCFAM 

will contain 10 sq. rd., 
for the quadrant ACDF 
is equal to the semicir- 
cle CAMC. 

(4) Now, taking away the com- 

mon segment X, we have 
left the triangle ACD = 
the lune CAMC. 

(5) As the triangle is equal to 

10 sq. rd., then the cres- 
cent contains 10 sq. rd. 




FIG. 55. 




(1) Let ABC be a right triangle, 

and describe a semicircle 
on each side. 

(2) The area of ABC equals the 

area of the two lunes Z and 
X taken together. 

(8) The area of the semicircle 
described on AB=the sum 
of the semicircles described FIG - S6 - 

upon AC and CB. 

(4) Now from each take the two segments cut off by AC and 
CB, and the triangle ABC remains of the larger semi- 
circle, and two lunes Z and X of the two smaller semi- 
circles. Q. E. D. 

PROBLEM 370. 

May and Ruth were cutting- out tickets for a recital, from a piece 
of cardboard, and the part remaining- was in the form of a segment of 
a circle, the height and base measur- 
ing exactly 2 in. and 12 in. The girls, 
being mathematically inclined, set to 
work to find the diameter of the circle 
from which the. segment was cut: 
what was the diameter? 

Solution. 

(1) Let ABC be the segment, 

and CD=2, AD=12 in. 

(2) Join CD, and produce CD to 

E, join EB. Then the two 




MENSURATION. 



179 



triangles CDB and BDE are similar. 

(3) We have CD : DB::DB : DE, or 2 : 6::6 : DE, or DE— 

18 in. 

(4) DE+DC = EC, or 20 in. 

.*. The diameter was 20 in. 



PROBLEM 371. 

Three equal circles touch each other externally, and thus enclose 
one acre of ground: what is the diameter in rods of each of these circles? 

Solution. 

(1) Draw three equal circles to 

touch each other externally 
and join the three centers 
ABC, thus forming an equi- 
lateral triangle. 

(2) Let R represent the radius of 

the three equal circles; 
then it is obvious that each 
side of the triangle is equal 
to 2R. 

(3) This triangle encloses the 

given area, and three equal 
sectors. As each sector is 
a third of two right angles, 
the three sectors are equal to a semicircle. 

(4) The area of a semicircle whose radius is R = 




FIG. 58. 



TrR 



(5) The area of the whole triangle is 



■R 2 



2 



160. 



(6) But the area of the equilateral triangle is R 2 j/8. (Fig. 38.) 
■R* 



(7) .\ R»V3 = 

(8) R*= 



2 
320 



f 160, or R*(2v/3— tt)=320. 
320 



2V3— 3.1416 0.3225 
(9) .'. R = 31.48+ rd. 



992.248. 



PROBLEM 372. 

Find the length of the longest and shortest chords that can be 
drawn through a point 6 in. from the center with a radius of 12 inches- 



180 



FAIRCHILD'S SOLUTION BOOK. 



Solution. 

(1) The longest possible chord through any 

point is the diameter drawn through 
it, and the shortest is one perpendic- 
ular to the diameter. 

(2) Therefore, AB is the longest, and KP 

the shortest chord. 

(3) FB=FO+OB = 18in. AF=6 in. 

(4) The products of the segments AF and 

FB = the product of KF and FP. 

(5) 6Xl8=i/i08"=10.39+ in., or \ of KP. 

(6) 2X10.39 = 20.78 in., the shortest chord. 




FIG. 59. 



PROBLEM 373. 

Two parallel chords in a circle, on opposite sides of the diameter, 
are each 20 ft. long-, and the perpendicular distance between them is 
8 ft.: what is the diameter of the circle? 

Solution. 

(1) Let ABCD be the circle, and DC, AB 

=20 ft., EF=8 ft., FO=4 ft., and 
DE = 10 ft. 

(2) From D draw DO, then OED is a right 

triangle. 




10.77, the radius of 
(4) .'. 10.77 ft. X 2=21.54 ft., the diameter. 



(3) DO=V(10 2 +4 2 ) 
the circle. 



FIG. 60. 



PROBLEM 374. 

Required the area of the segment, of which the chord is 50 ft., and 
the height 3 ft. 

Solution. 

(1) 50 ft. = the base of the segment, 3 ft.=the altitude. 

(2) 27 cu. ft. = cube of the height of the segment. 

(3) 100 ft. = twice the base. 27-M00=.27 sq. ft., the quotient 

of the cube of the height by twice the base. 

(4) 3x50 = 150 sq. ft. = product of the height and base. 

(5) \ of 150=100 sq. ft. = f of height and base. 

(6) 100 sq. ft.-h.27 sq. ft. = 100.27 sq. ft. 

:. 100.27 sq. ft. = area of the segment. 

Rule. — Divide the cube of the height by twice the base, and 
increase the quotient by two-thirds of the height and base. 



MENSURATION. 



181 



PROBLEM 375. 
E- F. Wieser cut off a segment of a circular cardboard, which had 
just \ of the circumference, and contained 130 sq. in.: what was the 
diameter of the cardboard? 

Solution. 

(1) CDB = ^ of the circumference. 

(2) BCD = the segment, angle COB = 120°, 

and angle COD=60°. 

(3) AO, or CO=r, the radius of the circle. 

(4) CF=-irV3, from Fig. 36. 

(5) The area of the sector COB=^r 2 7r, and 

the area of the triangle COB=CFX 
FO, or£r 2 V3. 

(6) .*. The area of the segment = CBD=£r 2 7r 

sq. in. 

(7) We find r=14.5 + in. 
.'. 29+ in. is the diameter of the cardboard. 




FIG. 61. 



)/' 2 V 3 = 130 



PROBLEM 376. 
From a point without a circle two tangents are drawn, forming 
with each other an angle of 60°, and the length of each tangent is 20 
in.: what is the diameter? 

Solution. 

(1) Let DB and AB, the tan- 

gents = 20 ft., and the an- 
gle ABD=60°. 

(2) The angle A is a right 

triangle because it is 
formed by a tangent and 
radius. 

(3) OB bisects the angle DBA; 

hence, angle ABO=30°, K1( 62 

and angle BOA=60°. 

(4) .*. The triangle BAO = half of an equilateral triangle 

whose sides are equal to OB. 

(5) OA=^OB. ABz=OAV3. 

20 

(6) .'. AOV3 = 20, or OA=— ==11.548+ in. 

(7) 11.548x2=23.096 in., the diameter. 




PROBLEM 377. 
A horse is tethered from one corner of a square 20-acre field: how 
long must the rope be so that it can graze over 20 acres outside the field? 



182 



FAIRCHILD'S SOLUTION BOOK. 



(1) 



(2) 



Solution. 

Let AOCB be the field, and O 
the point at which 



horse 



the 
is tied. 

Let R = OH, the radius or the 
length of the rope by which 
the horse is tied. 




FIG. 63. 



(3) DOF= J of a circle, or a quad- 

rant, but the horse grazes over 
£ of a circle, or FED. 

(4) As f of a circle = 20 A., then 

there must be 6J A. in the 
quadrant DOF. 

(5) 26J A.=area of the circle. 

(6) R 2 7r=area of the circle. 

(7) /. R 2 7r=26f A., or 4266f sq. rd.* 

(8) R=36.85 rd., the length of the rope. 

PROBLEM 378. 

A wheel 8 ft. in diameter is run by means of a belt from a wheel 1 
ft. in radius; the distance between their centers is 12 ft., and the belt 
is crossed between them, causing- them to revolve in opposite direc- 
tions: what is the length of the belt used? 

Solution. 

(1) OMandO'M' 
are parallel; 
O M P a n d 
O' M' P are 
similar tri- 
angles. 

(2) Hence, we 
have the pro- 
portion, O'M' 

: OM :: O'P 
: OP. 

(3) Now, PO'=*, OP=12-*. 

(4) 4:1::^: 12—.*, whence *=9.6 ft. 

(5) OP = 12 ft.— 9.6 ft.=2.4 ft. 

:8.72 ft. 

KP: 




FIG. 64. 



(6) PK'=: VPO' 2 — K'C 

(7) KP = Vop2_oK2 r=2 .18 ft. 

(8) KK'=KP+PK'=10.9ft. 

(9) Hence, MM'+KK'=21.8 ft. = length of the belt not on 
the wheels. 



MENSURA TION. 



183 



(10) K f S=(PK'xK'O f )-f-PO'=3.63ft. 

(11) K'M'= 3.63x 2=7 .26 ft. 

(12) SO'=VIW' 2 — KS 2 =1.68ft. 

(13) Area of segment K'ZM'=(ZS 3 -^by twice K'M') + 
l(ZSXK'M') =12.0886 sq. ft. 

(14) Area of triangle M'0'K'=*(K'M'XSO') =6.0984 sq. ft. 

(15) 12.0886 sq. ft.+6.0984 sq. ft.=18.186 sq. ft., area of sector 
O'K'ZM'. 

(16) 4 2 7r=area of large wheel, 87r=circumference of the large 
wheel. 

(17) The arc K'ZM'=*-4 2 : 18.186 ::8«- : (9 05 ft.) 

(18) 8tt— 9 05=16.08 ft = K'DM', part of circumference covered 
by the belt. 

(19) Area of the sector MLKO=4 2 : l 2 :: 18.186 : (1.13 sq.ft.) 

(20) 27r=circumference of small wheel. 

(21) The arc MLK=3.1416 : 1.13:: 6.2832 : (2.26 ft.) 

(22) 6.28 ft., circumference of small wheel— 2.26, arc MK= 
4.02 ft., part of small wheel covered by belt. 

(23) 4.02 ft + 16.08 ft. +21.18 ft =41.18 ft., required length. 

{Eaton.) 
If the belt is not crossed, what is its length? 

Solution, 

(1) Let F, L, S 

and K be the 
points where 
the belt 
touches the 
wheels. 

(2) LZ'O and 
LFD are sim- 
ilar triangles. 

(3) LD : LO::FD : Z'O; hence 3 : 4:: 12 : (16=OZ'.) 

(4) Z'0'=Z'0-0'0=4ft Z'F=VZ 7 Q' 2 — FO' 2 =3.87 ft. 




FIG. 65. 



(5) FL=VFD 2 — LU 2 = V12 2 — 3* = 11.618 ft. 

(6) 11.618 ft. X 2=23.236 ft, length of belt not touching the 
wheels. 

(7) Area of the triangle OLZ'=(Z'Lx LO)|=30.98 sq. ft. 

(8) LP= ( 30.98 X 2) +1 6=3.872 ft., LK=7.744 ft. 

(9) PO = VLO a — CF* =1,008 ft. 

(10) PZ = ZO— PO=2.997 ft. 

(11) Area of the segment LZK=(ZP 3 -^by twice LK) + 



184 



FAIR CHILD'S SOLUTION BOOK. 



f(ZPxLK)=16.488sq. ft. 

(12) Area of the triangle LKO=-J( LKx PO)=3.883 sq. ft. 

(13) Hence, the area of the sector OLKZ=16.488 sq. ft.+ 
3.883 sq.ft. =20.371 sq. ft. 

(14) 4*7r : 20.371 :: 8:r : (10.85 ft.)=arc LZK, or part that the 
belt does not touch of the large wheel. 

(15) 8tt— 10.85 ft.=14.28 ft., part of large wheel covered by 
the belt. 

(16) 42 : 12 :: 20.371 : 1.273 sq. ft., area of sector FMS, or 
part of small wheel covered by belt. 

(17) .5*-V : Itt :: 1 273 : .509 ft., length of belt on small wheel. 

(18) .509 ft.-f 23 236 ft.+14.28 ft =38.025 ft., required length. 

PROBLEM 379. 

A circular lot 15 rd. in diameter is to have three circular grass beds 
just touching each other and the larger boundary: what must be the 
distance between their centers, and how much ground is left in the tri- 
angular space about the center? ( Ray's Higher Arith.) 

Solution 

(1) OD = 7^rd. = R, the radius 

of the lot. 

(2) AO = 15rd._ 

(3) AB— AOV 3=25.98075 rd. 

CD=22| rd. 

(4) Area of ABC=^(ABx 

CD) =292.28343+ sq. rd. 

(5) The area of the three tri- 

angles AOB, COB and 
COA=J- of 292.28343 sq. 
rd.=97.42731 sq. rd. 

(6) Let SD, or SL=r, the 

radius of the small circle. 

(7) r=(AOBx2)-^AB+OB 

+AO, or twice the area FIG ' 66, 

of AOB divided by the perimeter=3.4807. 

(8) .\ SF=6.9614 rd. The area of sector PSL 

small circle, because the angle PSL=60°. 

(9) ^ 2 7r=3 times ^ 2 7r=area of the three sectors in the tri- 

angle KSF. 

(10) r 2 V3 — \r*ir=r 2 (y/3—iir) =1.9537 sq.rd., the area of the 
space enclosed. 

.*. SF = 6.9614 rd., and 1.9537 sq. rd.=area of the curvi- 
linear space. 




the 



MENSURATION. 



186 



PROBLEM 380. 

If 2 ft. be the radius of a circle, how far from the center must a 
chord be drawn parallel! to divide the semicircle into two equal parts? 

Solution. 

(1) Let ADB be the semicircle, AB 

the diameter, GH the line that 
divides the semicircle into two 
equal parts. 

(2) On OD, perpendicular to AB, let 

OF=;t, the distance required, 
and 0=angle HOB. r=OB, or 
radius. 

(3) Then *=rsin0, FH=rcos0. 

(4) Area of sector HOB=^- 2 0. 

(5) Area of sector DOH=|r 2 (^r— 0). 

(6) Area of triangle FOH=^r 2 sin0cos0. 

(7) We then have r 2 0-r-r 2 sin0cos0=r 2 (^r— 0) — r 2 sin0cos0. 




FIG. 67. 



(8) Transposing and reducing we have sin20-|-20 



2* 



(9) We will find the value of by double position, and it is 
found as follows: 



(10) 
(ID 



Let 0=2O o =<r for first trial. 



2tT TV 

=o« or 



9 



. sin40' 

.64279. 

(12) .-. .64279= .87267. 

(13) .87267— .64279=. 22988 
llir 
"90"- 



si n 40°=^-=. 87267, 
lo 



but sin40°= 



(14) Let0=22 c 



(15) 



11. 



(1) 



2& 



.-. Sin44+ — = -, or si n44°=^- =.80285. 
4o _ yu 



(16) .-. .69466=.80285. 

(17) .80285— .69466=. 10819 ... (2). 

(18) (1)— (2) =.12169, 22°— 20°=2°. 

(19) /.<#>: 2° :: .22988 : .12169. 

(20) .\ <£=3.77812 =3° 46' 47". 

(21) Sin47° 33'34"+47°33'34"=90°. 

(22) Sin47°33'34"=42°26'26"=|||^=. 74073. 

(23) . . .73798=74073. 



186 FAIRCHIL&S SOLUTION BOOK. 

(24) .74073— .73798 = .00275 . . . (3). 

(25) 23° 46' 47"— 22°=1° 46' 47"=. 6407". 

(26) (2) — (3) =.10544. 

(27) <f> : .6407" :: .10819 : .10544. 

(28) .*. <£=.6574"=1°49'34". 

(29) .\0=23°49'34", andsin47°39'8"=42°20'52"=?|^=.73911. 

(30) .-. .73907=.73911. 

(31) .73911— .73907 =.00004. 

(32) 23° 49' 34"— 23° 46' 47"=0° 2' 47"= 167". 

(33) .00275— .00004 =.00271. 

(34) /. <f> 2 : 167" :: .00275 : 00271. 

(35) .\ <£ 2 =169.5"=2'49.5". 

(36) 0=23° 46' 47"+2' 49.5". 

(37) 0=23° 49' 36.5". 

(38) Sin0=. 40397. 

(39) /. *=.40397r or .80794 ft. 

.*. .80794 ft., is the required distance. 

PROBLEM 381. 

What is the length of a chord cutting" off £ part of a circle, whoee 
diameter is 50 ft.? 

Solution. 

(1) From Fig. 67, the segment GDH is twice the segment 

ABHG, or 2r*0+2rZs'm0co±>0=r2 (±>n— 0)— r 2 sin0cob6L 

(2) Transposing and reducing, we have sin20-|-20=o- 

(3) Find the value of by double position as follows: 

(4) Let 0=14°==^ for first trial. 

(5) .-. Sin28°+^=^, or sin28°=^=. 55851. 

(6) But sin28°=.46947. 

(7) .\ .46947=. 55851. 

(8) .55851— .46947=.08904 ... (1). 

(9) Let 15°=^, and 30°+g=g. or 30°=^=.5235. 

(10) .-. .5000=.52359. and .52359— .5000=.02359 . . . (2)., 

(11) (1)— (2) =.06545, 15°— 14°=1°21'. 

(12) <f> : 1° :: 08904 : .06545. 

(13) .'. <^=1.3604 o r=l° 21' 38". 



MENSURATION. 



187 



(14) 14°+1° 21' 38"= 15° 21' 38". 

(15) .-. 0=15° 21' 38". 

(16) Let angle DOB=A=90°. 

(17) ... A_0=84° 38' 13"=angle DOH. 

(18) SinDOH=.96426+. 

(19) Multiply the diameter of the circle by .96426-j-, and the 

result will be the length of the chord. 

(20) . . .96426X50=48.21300 ft.=GH. 

PROBLEM 382. 
A circular field 80 rd. in diameter, is divided into three equal parts 
by two parallel fences: find length of fences, and the distance each is 
from the center. 

Solution. 

(1) Let AB and PF be the two fences, 

and KO, OE the distance from 
the center. 

(2) OB=r, angle BOT=0. 

(3) This is the same as the preceding 

equation for 0, i. e., sin20-{-20= 

21' 38". 




4 



FIG. 6K. 



k, or0=15 c 

o 

(4) Angle KOT=X=90°. 

(5) .-. A— 0=74°38'13"=KOB 

(6) Sin^= .96426+. 

(7) .'. AB=DT the diameter X.96426-J- =77.1408 rd. 

(8) KO=rsin0=lO.595 rd. 

.*. 77.1408 rd. is the length of the fence, and 10.595 rd. 
the distance from the center. 



PROBLEM 383. 
A circle containing- one acre is cut by another whose center is on 
the circumference of the given circle, and the area common to both is 
\ acre: find the radius of the cutting- circle. 

Solution. 

(1) Designate by O the center of the 
circle, and by A the center of the 
cutting circle. 
EA = R, the radius. 
With A as a center and radius R, 
draw the arc ECD, cutting the cir- 
cumference at E and D. 
(4) Join OD and AE. FIG . 69. 



(2) 
(3) 




188 FAIR CHILD'S SOLUTION BOOK. 

(5) Put r=OA = ±( — "W7.136 rd., radius of the 1st circle. 

(6) Put angle EAO = angle OEA=0. 

(7) Then angle EOA=tt— 20. 

(8) Sector EOA:= ^ -, sector EAC = — ~ — , triangle 

EAO=iRrsin0. 

/en tu u ^ 2 (tt-20) ,ttR 2 Trr 2 

(9) Then we have ~ -\ — 5 %Krsmd=— r . 

(10) But R=2rco&6. 

(11) Substituting the value of R, we have <jrr 2 (7r — 20) -j- 

4^2cos 2 0— wr 2 sin20=-£*rV 2 , or 20cos20— sin20=-|ir, or 
sin20— 20cos0=-i?r ... (1). 

(12) Let <£=20, and the equation becomes, sin<£ — <£cos£= 

*r • - . (2). 

(13) It is obvious that <£ cannot be less than \-x. Let us as- 

sume <f>=\Tr, and substitute this in (2). 

(14) Then sin<£ = l. ^ttcos<£=0. Now, 1 — 0=1, which is less 

than \tt. 

(15) Assume 4> = 135°=f*r, and sin</>=. 70711 

— f cos<fr — 1 66609 

2.37320, which is greater 
than \v. 

(16) Then, 2.37320 135° 1.57080=-^. 

L ' 90°_ L 

1.37320 : 45° :: .57080 : 18.7°, the correction to 
be added to the 1st assumed value of <[>=%*, giving 
108.7°. 

(17) Let us now assume 108° and 110° as the values of 4>. 

(18) For ^=108 o =|tt, we have sin<£= .95106 

— f?rcos4>— .58249 

1.53355, less than \*. 

(19) For <f>z=zll0°—\i<7r, we have sin<£= .93969 

—\lircos<f>= .65663 

1.59632, greater than i -JW. 

(20) Then, as before, 1.59632 110° 1.57080 

1.53355 108° 1.53355 
.06277 : 2° :: .03725 : 1° 11', which 
added to 108°=109° 11'. 



MENSURA TION. 



189 



(21) Now, take <£=109° 11' and 109° 12'. 

(22) For <£=109° ll r =^/Ws r > we have sin<£= -94447 

— tWA^os^ = .62616 
1.57063, 

less than -|?r. 

(23) For £=109° 12' = T Vo^ we have sin<£= .94438 

— T Vo7rcos^>= .62679 

1.57117, 

greater than \tt. 

(24) As before 1 57117 109° 12' 1.57080 

1.57063 109° 11' 1 57063 
.00054 : V :: .00017 : 18.8", which ad- 

ded to 109° 11' gives <£ = 109° 11' 18.8". 

(25) Dividing by 2, 0=54° 35' 39.4". 

(26) .'. R=2rcos0=8.26926 rd. 



PROBLEM 384. 

A horse is tied to a stake in the circumference of a 10-acre field: 
how long- must the line be to allow him to g^raze over just one acre in- 
side the field? 

Solution. 

(1) Let P be the point to 

which the horse is tied, 
B and A the points in 
the circumference of the 
field to which the horse 
can graze, and O the cen- 
ter of the field. 

(2) Let BO=r=V16(XH-7r = 

the radius of the field, 
and angle BPO = angle 
OBP=0. 

(3) Draw OD=rsin0 perpen- 

dicular to BP = 2PD = fig. 7,) - 

2rcos0, the length of the line. 

(4) Now BOP is isosceles, angle PBO = angle BPO, angle 

BOP=(tt— 20), and BP = R. 

(5) Sector BPE= TRi ^" 2tf) , sector BOP=^, A OAP = 




|R/-sin0. 
(6) Then we have 



7tR2(7T— 20) . Trr 



4Rrsin0 = ' 



rR2 

20 



(7) But r=2Rcos0. 




190 FAIRCHILD'S SOLUTION BOOK. 

(8) Substituting the value of r, we have *-R 2 (ir— 20)4- 

47r0R2 C os 2 0-^R2 s in20= 1 Vr R2 » or sin20— 20cos20= T Vr. 

(9) Solving this equation by the method of double position, 

we find p=76°21'45 f '. 
(10) BP=2rcosp= 10.6418 rd., the length of the line. 

PROBLEM 385. 

Find the area of the largest five-pointed star that can be cut from a 
circular piece of pasteboard 20 inches in diameter. 

Solution. 

(1) Draw the circle, and dividing the 

circumference into five equal parts, 
join the alternate points of division, 
as in the figure. 

(2) Then ABCDE will be the star in- 

scribed in the circle whose center 
is O. 

(3) The angle ECA = 36°, being meas- 

ured by half the arc EA. fig. 71. 

(4) Since OC bisects the angle, then angle FCO=18°, and 

and angle FOC=36°. 

(5) Hence, angle CFO = 180°— (36 -hl8°)=126°. 

(6) The sine of the angle opposite the given side is to the 

sine of the angle opposite the required side, as the 
given side is to the required side. 

(7) From this we have sinl26° : sinl8° :: 10 : (FO=3.832.) 

(8) Now, it is obvious that the triangle FOC is ^ of the 

star. 

(9) Multiplying half the product of the two sides by the 

sine of the included angle, gives the area of the triangle 
FOC, or AFOC=-i(10x3.832)Xsin36 o =11.2619 sq. in. 
(10) Area of the star is 10X11.2619=112.619 sq. in. 

NoTK. — This fig-ure was supposed to possess mysterious properties, 
and was called "Health." It was used as a badge by the secret society 
founded by Pythagoras about 550 B. C, for the pursuit of mathematics. 

PROBLEM 386. 

A deer is tied to a stake in the circumference of a 10-acre field: 
how long- must the line be to allow it to graze over just one acre out- 
aide the field? 

Solution. 

(1) From Fig. 70, let the larger circle be the field, P the 
point and PB the tethering line outside the field. 



MENSURATION. 191 

(2) Let OB=r, and X=angle BPO. 

(3) Then BP=2rcosX, arc BMA=2?r— 2X, and its length = 

2rcos\{27r—2\). 

(4) BOP=tt— 2X; then the area of BOP=^ 2 sin(7r— 2X) = 

-|r 2 sin2X. 

(5) Area of BOAP=r 2 sin2X; arc BP=-h— 2X. 

(6) Area of the sector BOP=^ 2 (7r— 2X) and sector O—BPA 

=r 2 (*— 2X). 

(7) .". The area of the portion of the circle outside is sector 

P— BMA+2ABOP— sector O— BPA=4r 2 cos 2 X(7r— X) 
-hr 2 sin2X— r 2 (ir— 2X), or cos2X(2tt— 2X)+sin2X=— fa. 

(8) Then, by position, we obtain the value of X. 

(9) LetX=70°. 

(10) /. 2(tt— ^^coslSO^sinlSO^— fa. 

(11) -.86603 x {*+%=- fa. 

(12) /. .5=.34677. 

(13) Let X=77°. 

(14) .\ — 1.028615tt+.43837=— , fa 

(15) /. .43837 = .40406. 

(16) .5— .34677 = . 15323 ) ^ooq 0341— 11RQ2 
.43837— .40406 \ - iD ^— .l*Wl— . n«yz. 

(17) /.2° :x°::. 11892 : .15323. 

(18) .*. ^°=2.2407°=2° 14' 26V- 

(19) .\ X=77° 14' 26V- 

(20) .-.— 1.030385H-.43080=.9ir. 

(21) /. .43080=. 40962. 

(22) .43080— .40962 = .02118. 

(23) .03431— .02118=.01313. 

(24) .2407 : x\\. 01313 : .03431. 

(25) .\*=.6898. 

(26) /. 77.61797°=77° 37' 44 3". 

(27) .*. — 1.0330377r+.41854=— .9ir. 

(28) .-. .41854=.41795. 

(29) .41954— .41795=. 00059. 

(30) .02118— .00059=. 02059. 

(31) 77.62898— 77 2407 = .38828. 

(32) .38828 : x\\ 02059 : .02118. 

(33) .-. *=.39944. 

(34) X=77.64014 o =77° 38' 24.5". 

(35) PB=2rcosX=9.66+ rd. 



192 



FAIRCHILD'S SOLUTION BOOK. 



PROBLEM 387. 

Three men own a grindstone 2 ft. 8 in. in diameter: how many 
inches must each grind off to get an equal share, allowing 6 in. waste 
for the aperture? (/?. //. A., p. 406, prod. 84.) 

Solution. 

(1) Let AT be separated into 

three equal parts, AH, 
HD and DT. 

(2) Let perpendiculars at the 

points of division be 
drawn to intersect the 
circumference construct- 
ed on AB as a diameter, 
and these points F and 
E be joined with B; they 
will be radii that will de- 
termine rings of equal 
area. 

(3) BT is the radius of the 

aperture, or 3 in., which subtracted from 16 in. leaves 
AT=13 in. 

(4) The area of the stone=16 2 7r=2567r. 

(5) The area of the aperture is 3 2 7r=97r. 

(6) The area to be ground off is 256tt — 97r = 2477r. 

(7) Each man's share will be ^ of 2477r=82|ir. 
2567T — 82j7r=173|7r, area of the stone after the first grinds. 




FIG. 72. 



(8) 
(9) 

(10) 

(11) 
(12) 
(13) 
(14) 



173|7r — 82j7r=91^7r, area of the stone after the second 
grinds. 

Dividing the area of a circle by tt and extracting the 
square root give the radius. 

AB = V "(256^T= 1 , 6 in. 

HB = V(73i»-Hr) =13.178 in. 

SB = V(173i7r-Hr):=9.557 in. 

.-. The first man grinds off 16-13.178=2.822 in; the sec- 
ond 13.178—9.557=3.621 in., and the third 9.557—3 = 
6.557 in. 



PROBLEM 388. 

Find the radius of the largest circle that can be drawn in a quad- 
rant of a circle, radius 20 in. 

Solution. 

(1) Bisect the given arc TB in S. 

(2) Let fall the perpendicular SF, join O with S, and produce 

it, making SP=SF. 



MENSURATION. 



193 



(3) Join P and F; draw SL 

parallel to PF, O'L to OT. 

(4) Then with the center O' 

and radius CS=r, de- 
scribe the circle SLE 
and it will be the in- 
scribed circle. 

(5) Since the triangles OSL 

and OPF are similar, and 

OF = OS,or20-^V2 = 
14.142 in.; then by simi- 
lar triangles, we have 
OP : OS:: OF : OL, or 
34.142 : 20 :: 14.142 : OL 
= 8.284 in. /. r=8.284 in. 




FIG. 




FIG. 74. 



PROBLEM 389. 
The radius of a circle is 6 in.; through a point 10 in. from the cen- 
ter, tangents are drawn: find the lengths of these tangents, and also 
the chord joining the points of contact. 

Solution. 

(1) Let O be the center of the 

given circle, and P the 
given point. 

(2) On OP as a diameter describe 

a circumference at B and C. 

(3) Draw BO and CO. 

(4) Angles OBP and OCP are 

right angles, being inscribed 
in a semicircle. 

(5) OP is 10 in,; OB, 6 in. 

(6) BP or CP, V(10 2 — 6 2 )=8 in. 

(7) Area of OBP is(8x6)-^2 = 24 sq. in., and its altitude BD 

= (24x2)-=-10=48in. 

(8) Hence, the chord is 96 in., and one of the tangents 8 in. 

PROBLEM 390. 
A horse is tethered to one corner of a barn 40 ft. square, by a rope 
100 ft. long: over how much ground can he graze? 

Solution. 

(1) Let ADCO be the barn, and OF the rope 100 ft. long. 

(2) The area of E?;F=f of 100 2 tt = 23562 sq. ft. 

(3) CE=AF = AS=CP = 60 ft. 

(4) Area of the two quadrants CEP and AFS=^ of 60 2 tt= 

5654.88 sq. ft. 






194 



FAIRCHILD'S SOLUTION BOOK. 



(5) 



HG = V(AG 2 - 
52.91 ft. 
DG=52.91-28.28 



AH 2 ) 




24.63ft. 
2 +2)=DL=17.41 



FIG. 



The area included in 
DKGL has been grazed 
twice, hence its area must 
be taken from the area 
already found. 

(6) DP=DS=20 ft. 

(7) HG = OD=40V2,or56.56. 

(8) AH=|OD = HD=28.28ft 

(9) AG=CG=6 ft. 
(10) 

(ID 

(12) V(24.63 2 +2)=DL = 17.41 ft., side of square KL, and its 

area is 17.41 2 =303.108 sq. ft. 

(13) The segments KSG and PLG are equal. 

(14) KS=LP=20 ft.— 17.41_ft.=2.59ft. 

(15) Area of segments is (KS 3 -4-KGx4) +f of (KSx KGx2) 

=60.372 sq. ft. 

(16) Area grazed twice is 303.108+60.372=363.48 sq. ft. 

(17) .'. The horse grazes over 23592+5654 88— 363.48=28853.4 

sq. ft. 

PROBLEM 391. 
A stake to which a horse is tethered, is 120 ft. from, the corner of a 
barn 60 ft. long-, 40 ft. wide, and in line with the long- side of the barn: 
if the rope is 120 ft. long-, over how much ground can the horse graze? 

Solution. 

Let FCEA repre- 
sent the barn. 

O represents the 
point at which the 
horse is tethered. 

OC = 20ft.,CE=40ft. 

OE 



(1) 
(2) 

(3) 
(4) 



20 2 ) 




v / (40 2 
= 44.721+ ft. 

(5) 44.721 : 20::sin90° : 

sin26° 34 r = an g 1 e 
CEO. 

(6) The large angle 

FXDO=270°+26° 34'=296H°- 

(7) 360° : 296H°::120 2 7 r : 37267.753 sq. ft., area of FODX. 

(8) (40x20) +2 = 400 sq. ft., area of triangle COE. 



MENSURA TION. 



195 



(9) Angle DET=90°— 26° 34'=63° 26'. 

(10) DE=120 ft.— 44.72 ft. =75.28 ft. 

(11) 360° : 63i|°::75.28 2 7T : 3137.07 sq. ft., area of TED. 

(12) MF=120 ft.— (60 ft.+20 ft.)=40 ft. 

(13) tt(40 2 -^4) =1256.64 sq. ft., area of MFA. 

(14) AT = 120 ft.— (60 ft. +44.75 ft.) =15.28 ft. 

(15) tt(15.28 2 ft.H-4) =182.374 sq. ft., area of ATH. 

(16) The area of AGH is common to the two sectors FMA 

and ATH, hence its area must be subtracted from the 
two. 

(17) Area of GAH = area of sector AGH + (area of sector 

FGA— area of triangle FGA). 

(18) In the triangle FGA, FA=40 rd., and AS = 7.64 rd. 

(19) 40 : 7.64::sin90° : sinll°. 

(20) Angle AFS = 11°. Angle FAS=90°— 11°=79°. 

(21) 360° : 79°::15.28 2 tt : 160.96 sq. ft., area of sector AGH. 

(22) 360° : 22°::40 2 tt : 307.61 sq. ft., area of sector FGA. 

(23) Then the area of the triangle FGA = 299.95 sq. ft. 

(24) 160.96+ (307.61— 299.95) =168.62 sq. ft., common part of 

both sectors. 

(25) 37267+400+3137.07+1256.64+182.374-168.62 = 42075.22 

sq. ft., part grazed over by the horse. 

PROBLEM 392. 

The distance over a mountain is 2 miles; the horizontal distance, 
through, is 1 mile: how many more pickets would be required to build 
a fence over than through, the pickets to be set perpendicular to the 
horizon in each case, supposing 1200 pickets are required to build the 
fence through the mountain? 

Solution. 



(1) It will take exactly the 

same number. The ac- 
companying cut will il- 
lustrate plainly. 

(2) AB = 1 mi.; ACB = 2 mi.; 

the perpendicular lines 
represent the pickets. 




FIG. 



196 



FAIRCHILD'S SOLUTION BOOK. 



V. ELLIPSE. 



93. An Ellipse is a plane curve of such a form that if from 
any point in it two straight lines be drawn to two given fixed 
points, the sum of these straight lines will always be the same. 

94. These two fixed points are called the foci. 

Practically, a tolerably accurate ellipse may be drawn on paper by 
sticking- two pins in it to represent the foci, putting- over these a bit of 
thread knotted tog-ether at the ends, inserting- a pencil in the loop, and 
pulling- the sheet tig-ht as the figure is described. 

The importance of the ellipse arises from the fact that the planets 
move in elliptical orbits, the sun being in one of the foci — a fact which 
Kepler was the first to discover. When we calmly reason upon the im- 
measurable distances, and the awful rapidity of motion, with the mass 
of matter moving in this beautiful curve, we are constrained to ac- 
knowledge that all our boasted knowledge is as nothing in the won- 
drous dispensation of Him who "telleth the number of the stars and 
calleth them all by their names." 

95. In Fig. 78, AA' is the major axis, or the diameter 
which passes through the foci. BB' is the minor axis, or the 
diameter which is perpendicular to the major axis. 

To Find the Area. — Multiply half the sum of the two diameters 
by it and the residt is the area. 

PROBLEM 393. 
If the axes of an ellipse are 60 and 80 ft., what are the areas of the 
two segments into which it is divided by a line perpendicular to the 
major axis, at the distance of 10 ft. from the center? 

Solution. 

(1) Let A'DAD' represent a cir- 

cle, EAS the segment of 
the ellipse A'BAB', and 
DAD' the segment of the 
circle. 

(2) CG=10 ft., DG2 = A'GxGA, 

or 27.000 ft. 



(3) DG = V1500, or 38.729 ft. 

(4) DD'=77.458 ft. 

(5) The area of DAD'= (h*+2b) 

+f of ^=1723.46 sq. ft. 

(6) Area of the segment of circle 

: the area of the segment 




FIG. 78. 



MENSURATION. 



197 



of the ellipse ::CB : CA'. 

(7) .'. 1723.46 : the segment of the ellipse:: 40 : 30. 

(8) From which we find the area of the segment of the ellipse 

to be 1292.56 sq. ft. 

(9) The area of the ellipse is (40 X 30) tt= 3769.92 sq. ft. (by 

the above rule). 3769.92—1292.56 = 2477.36 sq. ft., area 
of EAS. 

Note. — This problem was prepared by the author for the Teachers' 
Review. We wish to prepare an extensive work on the ellipse in the 
future. 

VI. CATENARIAN CURVE. 

96. A curve formed by a chain or rope of uniform density, 
hanging freely from any two points not in the same vertical 
line. 

The catenary was first observed by Galileo, who proposed it as the 
proper figure for an arch of equilibrium. He imagined it to be the 
same as the parabola. Its properties were first investigated by John 
Bernovilli, Huygens and Leibnitz. It is now universally adopted in 
suspension bridges. Each wire assumes its own catenary curve. 



PROBLEM 394. 

What length of rope is needed that if the ends are tied to two poles 
of the same height 100 ft. apart, the rope will sag 25 ft., midway be- 
tween the poles? 

Solution. 

(1) Let DE and PF be the 

poles; EF=100 ft. 

(2) Let D and P be the fixed 

points to which the ends 
of the rope are attached; 
the rope will rest in a 
vertical plane. 

(3) The curve DCP is called 

the common catenary. 

(4) Let C be the lowest point 

of the catenary. 

(5) Take this as the origin of 

coordinates, and let the horizontal line through C be 
taken for the axis of x, and the vertical line through C 
for the axis of y. 

(6) Let (x, y) be any point, P, in the curve. 




FIG. 79. 



198 FAIRCHIL&S SOLUTION BOOK. 

(7) Denote the length of the arc CP, by S. 

(8) Let c be the length of the rope whose weight is equal to 

the tension at C. 

(9) Now, if we move the origin to the point, O, at a distance 

equal to c below C, by putting y — c for y we have y= 

oK c ~f~^ c )» which is the equation of the catenary. 

(10) When x=o, y=c. 

(11) x varies from o to 50, and y from c to <:-}-25 at the same 

time. 

(12) 2S=the length of the rope. 

(13) To find c we have (*=2.7182818, the base of the Napier- 

ian Logarithms),— ^ z =--l+^+j-2+j : yg+ 1 2 3 4 ' etc * 

(14) ,-^1-^fi-^, etc. 

( 15) , z+ ,- z= 2+.,2 + _g_. 

(16) Now put* for*, and^^l+g^+gj^), etc - 

(17) Then when x—50 y y=c-\-25. 

/1fi^ tu _lok /-,, (50)2 (50)4 \ (50)* (50)* 

(18) Then c+2&=c{ l+*_-+^- f etc. J and ^- ? -+^— 

=25. Whence ^=53.71607. 

(19) rfS=-/^ 2 +^ 2 =^H-(^ a — * 2 ). 

(20) The limits of y of being £ and r-)-25, we have, S = 

V+25 ydy 

V (^+25) 2 — ^ 2 = V 50^+625 = 57.5356. 



f. 



\lyl— L 



(21) 2S = 115.0712 ft., or the length of the rope. 

VII. CYCLOID. 

97. A Cycloid is the curve which is produced when a circle 
rolls forward on a straight line. 

A familiar example of it is the bicycle wheel moving along- a 
smooth road. If a mark be made at any point on the circumference of 
a wheel, it will describe a series of cycloids. 

The curved figure thus produced is not, as the etymology suggests, 
"of the form of a circle"; were it so, then the point of the circumfer- 
ence commencing its revolutions at a given spot on the road, would, 
when that revolution was completed, return to that spot again. It does 



MENSURA TION. 



199 



not so return; but when having- completed, its revolution, it afresh 
touches the road, it is at an advanced point in it compared with the 
spot at which it before came into contact with it. 

98. Now, if the 
circle OP rolls on 
the straight line 
AX, the point P 
will describe the 
arc of a cycloid 
APHX. 

To find its equation, let A be the origin, O the center of 
the circle, r=OF, P = any point in the curve, and we have ^= 
AD, r=PD=LF. 




FIG. 80. 



But AF=arc PF^rvers" 1 J - . DF=PL=VBL . LF. 



/"vers" 



it 



equation. 

dx _ y 
dy~ \l% 



-DF = rvers" 



■lr 



■it 



■\j2ry—y' 1 , which is the 



.-. S=V2r f (2r—y) *dy = 
J o 

4r, which is \ the cycloidal arc. 



:ry — y 

( i i \2; 

(— 2(2r)2(2r— yY) 

The length of the branch AHX is eight times the radius, 
and the area of AHX is three times the area of the generat- 
ing circle. 

PROBLEM 395. 

Suppose a fly sits on the rim of a carriage wheel 5 ft. in diameter: 
what distance will the fly move while the wheel makes one revolution 
on a plane surface? 

Solution. 

(1) Let A, Fig. 80, be the initial position of the fly. 

(2) APH and X are subsequent positions as the wheel re- 

volves. 

(3) The fly describes the cycloidal curve APHX, whose base 

is the circumference of the wheel, or 5?r, 15.708 ft. = AX. 

(4) The branch AHX = 5x4, or 2^x8=20 ft. 

(5) The area of A HX = 5 2 ttX 8=58.905 sq. ft. 

PROBLEM 396. 

Suppose a fly lig-hts on the spoke of a carriage wheel 5 ft. in diam- 



200 FAIRCHILD'S SOLUTION BOOK. 

eter, 6 in. up from the ground: what distance will the fly move while 
the wheel makes one revolution on a level plane? 

Solution. 

(1) From Fig. 80, let P be the generating point at a distance 

from the centers P'0=£, A the origin, r=OF, 0=P'OF, 
*=AK,j/ = P'K. 

(2) Then x=rd — ^sin#, and y = r — bcosO, are the equations of 

the locus, in which b<j. 

(3) dy=bsin0d$. 

(4) dx—(r—cos$)dO, V (<fy) 2 + (d*) 2 = V (r*+b 2 — 2rdcosOdO. 

(5) Let S=the number of feet the fly moved while the wheel 

makes one revolution. 

(6) Then, since r=2-J and £ — 2, we have 

f "" V (r 2 +b*—2rbcos0de) = P" V (41— 4Ocos0d6). 
Jo J o 

(7) Put (9=2^; then S = 18 f? Vl— f£cos 2 <^ = 

Hl-H-3fe) 2 («) 2 -5«ife) 2 (I?) 3 -?(tWt) * («)*] 

= 18.84 ft. 

NOTE. — PNZ is a prolate cycloid, or the path described by the 
point P'; also the curve of the cycloid's origin would fall to the left of 
YA. I think the student will understand this; see Wood's Co-ordinate 
Geometry. 

VIII. CURVE OF PURSUIT. 

99. If a point moves along any path, and another point is 
made to move towards it according to any law, the path of 
the latter is called a Curve of Pursuit. 

PROBLEM 397. 
A fox 40 rd. due south of a hound is running- due east, 25 ft. per 
second; the hound gives chase at 
the rate of 30 ft. per second: how 
far will each run before the fox is 
caught? 

Solution. 

(1) It is evident that the 

hound is all the time 
running directly after 
the fox. 

(2) The curved line BC repre- 

sents the path of the 

hound. fig. 8i. 




MENSURA TION. 



201 



(3) Let P and F be their position at any instant, and P' and 

H their positions at the next instant. 

(4) Put AD=^, B?—s, AF=j, and PG = w. 

(5) Then EP'=4 PP' = ^, and PP'-HM=^, (A). 

(6) Now, let the rate of the hound be m times that of the 

fox. 

(7) Then mdy—ds. Put AB=b, AC = «. 

(8) Then from the triangles PP'E and HFM, we get dy : ds 

:: HM : dx. 
dydx 



(9) HM: 



ds 



(10) By placing for ds its value mdy, HM; 



dydx 



(11) From (A) we get, by substituting, ds- 



mdy 
dx 



dx 
m 



m 



dzv, or mdy 



: \-dzv, or m 2 dy 



dx-\-mdzv 



a) 



(2). 
and (2) becomes 



(12) Integrating (1), m 2 y=zx-\-mw-\-C . . 

(13) When x — o, y = o, and zv = b, C= — mb 

m 2 y = x-\-mzv — mb. 

(14) Now, m is a constant; if we can find its value for any 

time in the race, we can take it as the value of m. 

(15) Put x=-a and y = a, which is the case at the close of the 

race, and we get m 2 a — a — mb . . . (3). 

(16) From which a — — - — 7. 
v 7)l l — 1 

(17) In the given problem £ = 40 rd., and m=^. 

(18) Substituting these, we get ^ = 109^ rd., the fox's distance. 

(19) wtf=130}J rd., the hound's distance. 

(Prof. G. E. Kelly.) 

PROBLEM 398. 

A bird sits on the top of a pole 50 ft. high: how hig-h above the 
pole must a boy, standing- 100 ft. from the pole, aim his arrow, to hit 
the bird, the velocity of the arrow at starting- being- 100 ft. per second? 

Solution. 

(1) Let A represent the point where 
the arrow leaves the gun, D 
the foot of the tree, and E the 
point directly over the bird, 
where the arrow is aimed. 



(2) Let ^ = 50, the height of the tree, 
£=AD = 100, z>=100,z = initial 




FIG. 82. 



202 FAIR CHILD'S SOL UTION B O OK. 

velocity of arrow; /=time required for the arrow to 
make its flight to P, and ;r=EP, the distance required. 

(3) Then AE = ^/, the hypothenuse, and as x is the distance a 

body would fall in t seconds, we have t 2 —2x-r-g. 

(4) A\soAE 2 = v 2 t 2 =2v 2 x+g=(x+a) 2 +b 2 =x 2 +2ax+a 2 +b 2 

(5) Hence, x 2 + [2a —\ x =—(a 2 +b 2 ). 

v «*> * 

(6) x=^— a ±r(~aj — (a* +&*)!. 

(7) Substituting, x=ifipf£—50±[(i£Pf$—50) 2 — 2500+10000] 

=260.94527 ±235.78048=25.1648, or 496.7257 ft. 
Note;. — Solved by the author for the Teachers' Review. 

IX. PARALLELOGRAMS AND SOLIDS. 

100. A Parallelogram is a quadrilateral whose opposite sides 
are parallel. 

101. A Rectangle is a parallelogram whose angles are all 
right angles. 

102. A Rhombus is a quadrilateral whose sides are equal. 

103. A Square is a rectangle whose sides are all equal. 

PROBLEM 399. 

The diagonal of a square is equal to the side multiplied by ^/2. 

Solution. 

(1) Let ABCD be the square, and AC 

the diagonal. 

(2) AC 2 =AB 2 +CB 2 =2AB 2 , or 2CB 2 .* 



(3) AC=VAB 2 +CB 2 = V2AB 2 =ABV2 
.-. AC = ABV2. 




FIG. 3. 



PROBLEM 400. 
What is the side of a square whose diagonal is 20 ft.? 

Solution. 

(1) 20 ft.=diagonal. 

(2) 20 ft.-W2=i4.84 ft. = side of the square. 
Rule. — Divide the diagonal by the \/2. 



*This shows that the square of the diagonal of a square is equal to 
twice the square of one side, or twice the square itself. 



MENSURATION. 203 

PROBLEM 401. 
The area of a square is 800 sq. ft.: what is the diagonal? 

Solution. 

(1) 800 sq. ft. = area of the square. 

(2) 800x2 = 1600 sq. ft., double the area. 

(3) v 1600 = 40 ft., the diagonal. 

Rule. — Double the area and extract the square root. 

PROBLEM 402. 

What is the side of a square garden whose area is 3600 sq. ft.? 

Solution. 

(1) 3600 sq. ft. = area of the garden. 

(2) V 3600=60 ft., side of the garden. 

.*. 60 ft. = the side of the garden. 

PROBLEM 403. 

What is the area of a square whose diagonal is 10 ft.? 

Solution. 

(1) 10 ft. = the diagonal = AC. 

(2) 10X10 = 100 sq. ft. 

(3) 100 sq. ft.-i-2 = 50 sq. ft., the area of the square. 

Rule. — Divide the square of the diagonal by 2, the quotient will be 
the area. 

PROBLEM 404. 

The square of half a line is equal to \ the square of the whole line. 

Solution. 

(1) Let.s = AB. 

s 



(2) AP=^=| 

(3) AB*=.v*, APa = -^. 

.'. AP*=] of s*. 




FIG. 84. 

PROBLEM 405. 



If the square ABCD has 40 acres in it, find the distance from C to 
a point & the distance from B to A, as F. 



204 



FAIRCHILD'S SOLUTION BOOK. 



(1) 
(2) 
(3) 
(4) 



Solution. 
"BD 2 =40 acres, or 6400 sq. rd. 
AB = V 6400=80 rd. 



FB 



iof 80rd. = 16rd. 



Now, as FCB is a right-angled tri 

angle, FC= VCB2+"FB 2 = 

V80 2 +16 2 =81.5+rd. 
.-. CF=81 f 5+ rd. 

PROBLEM 406. 




FIG. 85. 



From a point in the side and 8 ch. from the corner of a square field 
containing- 40 A., a line is run, cutting- off 19| A.: how long is the line? 

[R. H. A.) 

Solution. 

(1) Let ABCD be the square 
field, and P the point 8 
ch. from D. 






(2) 



(3) 



(4) 



Draw PE parallel to DC, 
PF cutting off 19i A. 
above the line, or PFCD. 

DC, or DA = 20 ch.; then, 
the area of the rectangle 
DPEC=20x8 =160 sq. 
ch., or 16 A. 



The triangle EPF con- 
tains 19|-16=3| A. 
(5) EF, the side of the triangle=3^X2 







FIG. 86. 

A., or 70 sq. ch. 



(6) 

(7) 
(8) 

(9) 
(10) 



:20.3-f- ch. 
when the 19-| A. 



is taken above the line 



70-^-20=31 ch. 
PF=V20 2 +3£ 2 : 
.-. PF=20.3 ch. 

PF. 
Now, we will take it below the line PE. 
It is evident that the line PF will fall below its present 

position, as PF r shown in the figure. 

(11) Now, the quadrilateral PABF' contains 19| A. 

(12) DPEC+PABF=35| A. 

(13) PEF'=40 A.— 35i A.=4£ A. 

(14) Then, the right triangle PEF' = 4| A. 

(15) 4|X2 = 9 A., or 90 sq. ch., area of the rectangle PEF'S. 

(16) EE=90-^20 = 4ich.; that is, the area of a rectangle di- 

vided by its length gives its width, or vice versa. 



MENSURA TION. 



205 



(17) PF'=V20 2 +4£ 2-20.5 ch. 
.-. PF'=20.5 ch. 

PROBLEM 407. 

In a square field which contains 10 A., a line is drawn from the 
southwest corner to the middle of the north side; from the southeast 
corner to the middle of the north side; and from the southwest corner 
to the northeast corner: find the area of the triangle enclosed by the 
three lines. 

Solution. 



(1) 

(2) 



(3) 



Let ADEF be the field. 
The side DE is easily 



found to be 



is 
40 



rd., or 



(4) 

(5) 

(6) 
(7) 




V (10X160). 
The area of the triangle 

BEF is 20x40^2, or 400 

sq. rd., which is also the 

area of the triangle ABE, 

for it has the same base 

BE and the same alti- 
tude EF. 
Subtract the area of the 

triangle BEC from this, 

and the result is the area 

required. 
Since BE is half AF, the altitude of the triangle BCE 

half the altitude of the similar triangle ACF, for 
BE : AF :: BC : CF :: EC : AC :: GC : CH. 
.*. The altitude of the triangle is equal to \ of 40, or 13^ 

rd., and its area is 20X13J-J- 
Hence, 400— 133J = 266f sq. rd 

ABC. 



FIG. 



IS 



or 1334 
the 



5 sq. rd. 
area of the 



triangle 



PROBLEM 408. 



A man takes for himself a central square of 40 acres, within a 
square of 160 acres, and gives to his four sons rectangular farms equal 
in every respect and making up the remainder of the farm. The father 
and sons are to have dwelling's centrally located in their respective 
farms: how far will each son be from his nearest brother, and from his 
father? 

Solution. 

(1) Let ABCD be the farm of 160 A., and LFXM the fath- 

er's farm. 

(2) Draw EL, MP, XN, FY, and ARXY, RBPM, PCEL and 

EDYF will be the sons' farms. 



206 



FAIRCHILD'S SOLUTION BOOK. 



(3) 
(4) 
(5) 
(6) 
(?) 

(8) 

(9) 



(10) 

(ID 
(12) 
(13) 



AB = BC=160rd. 

LF=FX=80rd. 

YF = EL = MP=40rd. 

YD=AR=120 rd. 

OK = ON+NK=40 rd.-h 

20 rd.=60 rd. 
S'K=iAR— NX=60 rd.— 

40rd. = 20 rd. 



Then, S'0 = V (60 2 +20 2 ) 
= 63.24 rd. = the distance 
each son's house is from 
the father's. 

ST=60 rd., and TZ=20 rd. 

/. SZ=60 rd.+20 rd. = 80 rd. 

ZS'=40 rd. 

Then, SS'=-i/(40 2 +80 2 )=8c 
son is from the other. 




FIG. 88. 



.44+ rd.=the distance one 



PROBLEM 409. 

John has a square wheat field containing- 10 A., and hires Mart to 
cut it; he starts to cut at the southeast corner of the field: how far is 
the machine from the southeast corner of the square after cutting- 5 A., 
the machine being a 6 ft. cut? 

Solution. 

(1) Let ABCD be the wheat 

field, and EFGH the 
part that remains after 
cutting an integral num- 
ber of rounds. 

(2) KLMN=5 A., the part 

that remains providing 
the machine cuts the first 
5 A. in an integral num- 
ber of rounds. 

(3) AD=AB = 40rd. = 660 ft. 

(4) NK=KL=28.28 rd.= 

466.62 ft. fig. 89. 

(5) FP=40rd.— 28.28 rd. = 11.72 rd.-^2=5.86 rd.=96.69 ft. 

(6) 96.69-^6, the width of the machine = 16, or the integral 

number of rounds, and .49 rd. remaining. 

(7) .49 rd.=w, 1.38 ft., twice the width. 

(8) EF=KL+twice w=468 ft. 

(9) Area of E FGH =468 2 =219024 sq. ft., and area of KLMN 




MENSURA TION. 



207 



= 5 A., or 217800 sq. ft. 

(10) The area of the part that is not cut, to have a remainder 

of 5 A., is hh, or 219024—217800 = 1224 sq. ft. 

(11) Now this 1224 sq. ft. is to be cut by taking a full swath, 

at the southeast corner of EFGH. 

(12) XFST represents the swath containing 1224 sq. ft. 

(13) Now, we have the breadth of this rectangle (6 ft.); we 

find its length by dividing by its breadth, or XT = SF = 
1224^-6 = 204 ft. 

.*. The machine is 204 ft. from the southeast corner of 
FGHE. 

Note. — The latter part of the problem has reference to the square 
EFGH. 



PROBLEM 410. 

Through a man's farm of 1000 acres, lying- in the form of a square, 
runs a railroad in a straight line on the diagonal: what does the right 
of way cost at $200 an acre, the strip being 100 ft. wide? 

Solution. 

(1) Let ABCD be the farm, 
and FGCHEA 
strip occupied 

railroad. 

(2) The diagonal AC= 



(3) 



be 
by 



the 
the 




!/(1000X 160x2) = 
565.68551 rd., or 
9333.81091 ft. 

EAF and HGC are isos- 
celes triangles. 

HC=CG=AF=EA=50ft. 

(5) If we take a rectangle 100 ft 

wide and 9333.81091 ft. 

long, we have the area of the rail road + 4 smal 

isosceles right triangles. 

(6) The area of all is 50 2 X2=5000 sq. ft. 

(7) [(9333.81091x100)— 5000 ]-^43560, the number of square 

feet in 1 A. =21.3123 A. 

(8) 21.3123 X 200=14262.46, the cost of right of way. 



FIG. 00. 



equal 



PROBLEM 411. 

By cutting off a strip 3 in. wide from around a square board, I 
found that $ remained: find the area of the board. 



208 



FAIRCHILD'S SOLUTION BOOK, 
Solution. 



(1) Let ABCD be the board. 

(2) Let MNFE be f of the 

whole area. 

(3) MN = Vl=* of the side 

AB. 

(4) AB = f, then f— \-=\, twice 

NT. 

(5) NT=iof*=i. 

(6) NT=3in. 

(7) £=3. f, or AB=6x3=18 

in. 

(8) AB2— 182 in. = 324 sq. in., 

area of ABCD. 




FIG. 91. 



PROBLEM 412. 

Required the length of a piece of carpet that is a yard wide, with 
square ends, that can be placed diagonally in a room 40 ft. long and 30 
ft. wide, the corners of the carpet just touching the walls. 

Solution. 

(1) Let ABCD be the room, 

and GFEH the strip of 
carpet. 

(2) Let *=EC=AG, and y= 

CF = AH. 

(3) Then, by similar triangles, 

40— x : 30— y :: y : x. 

(4) Or, 40*— * 2 = 30?/— y* . (1). 

(5) We also have * 2 +i/ 2 =3 2 , 

or*=V(9-: y 2 ) • • (2), 

(6) Substituting in (1) and reducing, y±— 30j' 3 +616> 2 +135jj/ 

—3579.75=0. . . . (3). 

(7) /=2.4337, *= 1.7541. 

(8) Hence, the length is V (40— x) a + (30— ^)' a =47.144+ ft. 

Note. — This solution was prepared by the author forthe^m<?r*Va« 
Mathematical Monthly. 




PROBLEM 413. 

What is the area of a path 1 yard wide, passing diagonally across 
a lot 20 rd. wide and 30 rd. long, the diagonal of the lot passing 
through the middle of the path? 



MENSURA TION. 



209 




FIG. 93. 



Solution. 

(1) Let ABCD represent the 

lot; HGLE, the path. 

(2) EI, HS, GF and LK are 

each perpendicular to 
DB, and are each 18 in. 

(3) By similar triangles, 

DC : CB :: DS : HS, 
and DS = 2.25 ft. 

(4) DC :CB::GF: FB, from 

which FB— 1 ft. 

(5) The total area of the four triangles is 4.875 sq. ft. 

(6) DC 2 + BC* = DB 2 , from this we find DB=593.23+ ft. 

(7) 593 23 ft.— 3 25 ft. (DS+1) =589.98+ ft., SFor IK. 

(8) 589.98x3 ft. =1769.94+ sq. ft., area of the two rectangles. 

(9) 1769.94 sq. ft. + 4.875 sq. ft. = 1774.815 sq. ft., total area. 

PROBLEM 414. 

Find the diagonal of a rectangular field containing lHi A., who.se 
length is to its breadth as 12 to 5. 

Solution. 







fk;. *>4. 



(1) Let A BCD 

be the field. 

(2) Its area is 
13J- X 160 = 
2160 sq. rd. 

(3) Assume 
A'B'C'D' as 
a similar field 
whose sides 
are D'A' = 5, A'B'=lL\ 

(4) Then, similar figures are to each other as the squares of 
their like dimensions. 

(5) Area of A'B'C'D'=5X 12=60 sq. rd. 

(6) We have A'B'C'D' : ABCD : A^' 2 : AD 2 , or 
60 : 2160 ::5 2 : AD 2 . 

(7) From this we find AD— 30 rd. 

(8) A'B'C'D' : ABCD:;D'C' 2 : DC 2 , or 60 : 2160:: 12* : DC 2 . 

(9) From t his DC=72 rd . 

(10) AC = V(AB* + BO)=:7tf rd. 
.-. The diagonal AC = 78 rd. 



210 



FAIR CHILD'S SOLUTION BOOK. 



PROBLEM 415. 

The diagonal of a rectangle is 109 and its perimeter 302; required, 
the sides. 

Solution. 

(1) The diagonal of a rectangle divides it into two equal right- 

angled triangles. 

(2) Let x represent the perpendicular, y the base and h the 

hypothenuse of one of the triangles. 
{3) Then, we have x-\-y—lbl ... (1). 

(4) ^ 2 +^ 2 ^109 2 ... (2). 

(5) Subtracting (2) from the square of (1), we have 2xy= 

10920 ... (3). 

(6) Subtracting (3) from (2), we have x 2 —2xy+y 2 =961. 

(7) Extracting the square root, we have x—y—Z\ ... (4). 

(8) From (1) and (4), we find *r=91 and j/=60. 
{9) Hence 91 and 60 are the sides of the rectangle. 

PROBLEM 436. 

A corn field is 25 rd. longer than wide and contains 900 sq. rd.: 
what are its dimensions? 

Solution. 

(1) Let x represent the width, and jr-{-25 the length. 

(2) *(*+25),or*2+25;r=900. 

(3) Completing the square and extracting the square root, we 

find *=20 rd., and ;r+25=r=45 rd. 



PROBLEM 417. 

A rectangular flower garden 60 yd. long and 40 yd. wide, has a walk 
6 ft. wide around it, and paths of the same width through it, joining 
the middle points of the opposite sides: find in square yards the area 
of one of the four flower beds enclosed by paths. 

Solution. 

(1) Let ABCD be the garden. 

(2) Theareais60x40=2400sq.yd. 

(3) Let S, S', S", S'", be the flower 

beds. 

(4) The length of the side walk 

= 60 yd. X 2+36 yd.x2 = 
192 yd. 

(5) The length of the inside 

walks is 56 yd. -{-34 yd.= 
90 yd. 

(6) 192 yd+90 yd. =282 yd., the length of the walks. 




FIG. 95. 



MENSURATION. 211 

(7) Now, the area of the walks is 282 X 2=564 sq. yd. 

(8) The area of the flower beds is 2400—564=1836 sq. yd. 

(9) The area of one of the beds is 1836 sq. yd. +4=459 sq. yd. 

PROBLEM 418. 

At three corners of a square garden, each side 100 ft., stand towers 
20 ft., 60 ft., and 80 ft. high respectively: where must a ladder be placed 
in this garden, that it may reach the top of each tower without moving 
its base? 

Solution. 

(1) We must place the foot of the ladder in the line perpen- 

dicular to the side between the 20 ft. and 60 ft. towers, 
ata point from which the ladder will reach both towers, 
and, also, in the perpendicular to the second side at the 
point where the ladder would reach the tops of the 
60 ft. and 80 ft. towers. 

(2) Now, to find the first point, we have 20 2 -f;r a =60 a -f-> 2 . 

(3) * a +^=100; then * = 66,j = 34. 

(4) To find the second point, we have 60 a +# 2 =80 a +# 2 . 

(5) a- H?=100; then, ?< = 64, and ,g=3 6. 

(6) V (80H-- 2 +7 2 ) = V (60 a +* a +jp a ) = V (20 a +« a +* a ) = 

78852 = 94.08506+ ft., the length of the ladder. 



(7) From the first tower, the foot of the ladder is x/8852— 20 2 

= 91.934759+ ft. 

(8) From the second it is V8852+60 2 = 7i , .470rt21 + ft. 

(9) From the third it is V8852—80 2 =49.5176736 + ft. 

(10) The height of a tower on the fourth corner, that the 
ladder will reach is V8852— (j =40^% or 

56.5668+ ft. 

PROBLEM 419. 

A rectangular field is 64 rd. long and 50 rd. wide: if from the mid- 
dle of either side we set out a right line to its opposite, where must it 
intersect in order to cut the field in 
the ratio of 1 to 2? 

Solution. 

(1) Let ABCD be the field. 

and P the point of the 
side AH. 

(2) Draw the line PE perpen- 

dicular from P to the op- 
posite side. 

(3) Draw PF which shall di- 

vide the field as required. fic 




212 



FAIRCHILD'S SOLUTION BOOK. 



(4) The area of the field is 64X50, or 3200 sq. id. 

(5) APED=£ of the field, and PBCF=| of the field. 

(6) The triangle EPF=|— J=* of the field, or \ of 3200 sq. 

rd.=533jsq. rd. 

(7) Since DA=50 rd., EF=(533£X2)-^50=21£rd. 

(8) EC=32 rd., half the side. 

(9) FC=CE=FE = 32-21i=10f rd., the distance from the 

corner of the field. 




FIG. 97. 



PROBLEM 420. 

There is a square field such that the number of rods around it is 
equal to the number of acres within it: how many acres does it con- 
tain? 

Solution. 

(1) Let ABCD be the square field. 

(2) O is the center of the field. 

(3) Draw OF perpendicular to AB. 

(4) FE=1 rd. 

(5) Join OE. Then EFO is a right 

triangle. 

(6) Now, as EF=1 rd., the triangle 

EFO = l A., or 160 sq. rd. 

(7) OF=(160x2)-=-l=320sq. rd. 

(8) Then CB = 640 rd., the side of the 

square. 

(9) 640 rd.X 4=2560 rd., the perimeter. 

.*. There are 2560 A. in the field. 

PROBLEM 421. 

If 16 rails fence a rod, how many acres in that square field of such 
extent that every rail will fence an acre? 

Solution 

(1) Let the above figure represent the square. 

(2) EF = 1 rd. and the triangle EFO = 16 A., or 2560 sq. rd. 

(3) -OF= (2560x2)^-1=5120 rd. 

(4) CB = 5120 rd. X 2 = 10240 rd. 

(5) The perimeter of the field = 10240 rd.X 4=40960 rd. 

(6) Now as 16 rails fence a rod, and every rail will fence a rod; 

there will be as many acres in the field ABCD as the 
product of 40960x16=655360 A. 

Note. — The above problem has bothered many an applicant at the 
county examination. Now, my friend, do not miss it again. 









MENSURA TION. 



213 



PROBLEM 422. 

You have a square farm of 40 acres: find the side of that farm 
without using square root. 

Solution. 



(1) In the diagram, let ABCD = 1 sq. mi. 

=640 A. 

(2) Then EFGD= T V sq. mi. = 40 A., the 

required farm. 

(3) But ED=lmi.=80rd. 



PROBLEM 423. 

At the northwest corner of a rectangular field two men start to 
walk at the same rate, one east on the short boundary line and the other 
on the diagonal: where will they meet if the one turns at the south- 
east corner to meet the other, the field being 96 rd. long and 28 rd. 
wide? 




Solution. 



(i) 

(2) 

(3) 
(4) 

(5) 



From Fig 



94, let ABCD be the rectangle, AC the diago- 
nal, and P the point of meeting. 

V AD a +DU2=100 rd. = AC. 

AD+DC=124 rd. 
From AD to P=112 rd. 

PC=[(96+28)—100] = 12 rd., the point of meeting north 
of the southeast coiner. 

PROBLEM 424. 

Schwarts has a garden in the form of a square; from the corners to 
a spring within the garden, the distances are a=40, £=50, r=80 ft.: 
find the side of the garden. 

Solution. 

(1) Let ABCD be the square and F the spring. 

(2) DF = 50 ft., CF=40 ft. and FB = 80 ft. 

(3) Draw DE=50, EC = 40, CT=40, and TB=80 ft. 

(4) Suppose S is a spring whose distances are AS=40, DS = 

50, and SB = 80 ft. 

(5) VSB~*+BT* = V80 2 +80*= 113.13 ft.=ST. 



(6) SE=VSD a +DE2, orV50 2 +50 2 =70.71 ft. 



214 



FAIRCHILD'S SOLUTION BOOK. 



(7) 



(8) 



(9) 



(10) 

(ID 
(12) 
(13) 

(14) 



In the scalene tri- 
angle STE, EK is 
its perpendicular, 
and now we will 
find SK and KT by 
a well known theo- 
rem in geometry. 

TS : TE + ES:: TE 
— ES : KT— SK, or 
113.13: 150.71:: 9.29 
: 12.37. 

Half the difference 
of the segments ad- 
ded to half their 
sum gives the great- 
er segment, and subtracted gives the lesser segment. 

Therefore. KT is 62.75 ft., and SK=45.38 ft. 




FIG. 99. 



KE = VKT 2 — ET 2 =49.6 ft. 

Area of STE^jSTx EK = 2805.376 sq. ft. 

The area of SBT = 3200 sq. ft, and area of EDS = 12500 

sq. ft. 
Therefore, the area of DSBTE = area of ABCD = 

7255.375 sq. ft. 
.*. AB, the side of the square garden = 85-f- ft. 

PROBLEM 425. 
A square farm contains 40 acres. It is required to lay off another 
park containing - the same area, enclosed by an iron fence forming- cir- 
cular arcs only, and to find the cost of such a fence at $2 per rd. 

Solution 

(1) Let ABCD represent 

the square farm, whose 
si des AB, BC= 
V 40X160=80 rd. 

(2) With a radius equal to 

that of the circum- 
scribing circle draw 
the arcs APB and 
BRC. 

(3) The area of the pelicoid 

APRCD is equal to 
that of the square 
farm, and equal to the 
area of the park, 
bounded by circular 
arcs only. 







FIG. 100. 



MENSURA TWN. 



215 



(4) The fence required is the length of the circumference of 

the circumscribing circle. 

(5) 80x^=251.328 rd., the circumference of the circle, or the 

length of the fence. 

(6) .-. The cost is 251.328 X $2=$502.656. 

PROBLEM 426. 

What is the length of each side of an octagon formed from a square 
whose sides are 8 ft.? 

Solution. 

(1) Take A, B, C, D as centers. 

(2) With a radius equal to OA, 

aescribe arcs cutting the 
sides. 

(3) Join the adjacent points, 

and the resulting figure 
NMLKFGQP is the re- 
quired octagon. 

(4) OB=4V2, the diagonal of 

the square whose side is 
4 in. 

(5) The area of the octagon is 

the square less the area 
of four triangles cut off. 

(6) Then 8— 4^/2=2.3431456 ft., side of the triangle FBG. 

(7) /. 8— (2.3432456x2) =3.3133088 ft., side of the octagon. 

PROBLEM 427. 

The diagonal of a square circumscribed about a circle is 40 rd.: 
find area of inscribed square. 

Solution . 

(1) From Fig. 100, let EFGH be the circumscribed square, 

and ABCD the inscribed square. 

(2) EH, the diagonal of the large square =40 rd. 

(3) Area of the large square is 40 2 ^-2 = 800 sq. rd. 

(4) Let OC—r, then EFHG = 4r 2 , and ABCD=2r~. That is, 

the area of the large square is twice the area of the 
small square. 

(5) /. The area of ABCD = V of 800 sq. rd.=400 sq. rd. 

PROBLEM 42s. 
To make a square equal to a g-iven rectangle ABCD. 




FIG. mi. 



216 



FAIRCHILD'S SOLUTION BOOK. 






Solution. 

(1) Produce one side AB, till 

BE is equal to the shorter 
side of the rectangle 
ABCD, or CB. 

(2) On A E as a diameter de- 

scribe a circle meeting 
BC produced at F. 

(3) Then will BE be the side 

of the square BFGH, 

equal to the given rectangle DB, as required. 

(4) AB : BF^:BF : BE. 

(5) Hence BF*=AB.BE, or AB.BC. 




FIG. 102. 



PROBLEM 429. 

On a rectangular lot whose sides are 39 ft. and 49 ft., a house is 
built, covering- 999 sq. ft.; the house is so located and of such a shape, 
that the lot surrounding- the house is of uniform width: find the basal 
dimensions of the house. [Helfrich.) 

Solution. 



(i) 

(2) 
(3) 







Let ABCD represent the rectangular lot and SK the 

house. 
Divide the lot into four equal parts as represented in Fig. 

103, and arrange these four figures as shown in Fig. 104, 

which makes the square HN. 
PC=24| ft., and CR = 19£ ft. 

(4) Then, the side of the square HN = 19^4-24^=44 ft. 

(5) The area of HP=442=1936 sq. ft. 






MEN SURA TION. 21 7 

(6) The area of the lot DB = 49x 39 = 1911 sq. ft. 

(7) The area of the lot surrounding the house is 1911 — 999 = 

912 sq. ft. 

(8) 1936—9 12 = 1 024 sq. ft., area of the square DE. 

(9) DF = V1024 = 32ft. 

(10) (44—32)^-2 = 6 ft. = ZX, the width of the lot. 

(11) ZX = TK. From this we can see that the width of the 

house is 39 ft.— 12 ft .=27 ft., and the length 49 ft.— 12 
ft. =37 ft. 

PROBLEM 430. 

My lot contains 135 sq. rd., and the breadth is to the length as 3 to 
5: what is the width of a road which shall extend from one corner half 
around the lot and occupy \ of the ground? (R. H. A.) 

Solution. 

(1) From Figs. 103 and 104, let AC represent the lot con- 

taining 135 sq. rd., and SK a similar lot containing 15 
sq. rd., and whose sides are 3 and 5 rd. 

(2) Then by similar figures, 135 sq. rd. : 15 sq.rd. :: DA 2 : VS 2 , 

or 135 sq. rd. : 15 sq. rd. :: DA 2 : VS 2 , from which DA 
=9 rd. 

(3) 135 sq. rd. : 15 sq. rd. :: AB 2 : VK 2 , from which AB= 

15 rd. 

(4) Now let the rectangle IX in Fig. 104 represent AC in 

Fig. 103. , 

(5) Then construct the rectangles IS'FQ, FOMPand MJXN. 

(6) They are equal to IHXH, for their lengths and breadths 

are equal. 

(7) HN=9 ft. +15 ft. = 24 ft. 

(8) The area of HP=24 2 =576 sq. rd. 

(9) The area of the road is 135 sq. rd. 

(10) The area of DE=576— 135 = 441 sq. rd. 

(11) FD, or FE= V441=21 rd. 

(12) 24 rd.— 21 rd. = 3 rd. 

(13) ZX = | of 3=1£ rd. 

(14) \\ rd.=24| ft., the width of the road. 

PROBLEM 431. 
How wide a strip must be plowed around a square 40-acre held to 
plow 75% of it? 

Solution. 

(1) The field contains 6400 sq. rd., and the side is 80 rd. 

(2) Since 75%, or \ of it is plowed, the square unplowed is \ 



218 



FAIR CHILD'S SOLUTION BOOK. 



of the field. 

(3) Vi. or i °f {t is i ° f the side of the field, or 40 rd. 

(4) Hence the width of the strip plowed is (80— 40) +2 = 20 rd. 



PROBLEM 432. 

The length and breadth of a ceiling- are 6 and 5; if each dimension 
were 1 ft. longer, the area would be 304 sq. ft.: what are the dimen- 
sions? (R. H. A.) 

Solution. 





FIG. 105. 



FIG. 106. 



(1) 304 sq. ft.— 1 sq. ft.— 303 sq. ft. 

(2) 6x5 = 300 squares as shown in Fig. 105. 

(3) 6+5=11 rectangles 1 ft. wide. 

(4) 30 squares+11 rectangles = 303 sq. ft. 

(5) 303+30=10^0 sq. ft., area of each square. 

(6) 11 rectangles +30=4^ sq. ft., addition to each square. 

(7) i-i+2=fL ft., width of the small rectangle in Fig. 106. 

(8) (U) 2 =-ii(h S Q- ft -» area of square O in Fig. 106. 

(9) W -=WoV- sq.ft. 

(10) -^VoV+aVV^WA 1 sq. ft. 

(11) Extracting the square root of this fraction, we have J^ 

ft., side of square enlarged, Fig. 106. 

(12) ^ ft. = the addition. 

(13) VV — H=W-=^ ft 'ri side of squares in Fig. 105, room 5 

by 6 squares. 

(14) 5X3=15 ft, one dimension. 

(15) 6x3 = 18 ft., the other dimension. 

Note. — Solved for the Teachers' Review bv N. D. Moser. 



PROBLEM 433. 

What is the area of a trapezium the diagonal of which is 110 ft. 
and the perpendiculars to the diagonal 40 and 60 ft. respectively? 



MENSURA TION. 



!19 




FIG. 107. 



Solution. 

(1) Let ABCD be the trapez- 

ium, AC the diagonal. 

(2) DE=40ft., and FB = 60 ft., 

the perpendiculars to AC 

(3) 110 ft. = AC=base of the 

triangle ACD. 

(4) ^(ACxDE)=2200 sq. ft., 

area of the triangle. 

(5) AC=the ba^-e of the tri- 

angle ACB. 

(6) FB=altitude of ACB. 

(7) .'. \ of (ACX FB)=330O sq. ft. = area of ACB. 

(8) Then ACD + ACB=5500 sq. ft., area of trapezium. 

PROBLEM 434. 

Find the perimeter of a rhombus, area 21(j sq. ft., and one diagonal 
24 ft. 

Solution. 

(1) Let ABCD be the rhombus. 

(2) AC=24 ft., the given diag- 

onal, and DB is the other 
diagonal. 

(3) The triangle DBC=| of 216, 

or 108 sq. ft. 

(4) The area of the right triangle 

DFC=* of 108=54 sq. ft. 

(5) DF=(54x2) — FC, or 12= 

9 ft. 

(6) Then DB = 9x2=18 ft. 

(7) DC= VDF* + FC* = 15 ft., the side of the rhombus. 

(8) .'. The perimeter is 15x4, or 60 ft. 




PIG. 108. 



PROBLEM 435. 

The area of a rhombus is 96 sq. rd., and the diagonals are to each 
other as 3 to 4: find the side. 

Solution. 

(1) Assume a similar rhombus whose sides are A'B', B'C, CD' 

and DA', diagonals A'C and D'B'. 

(2) F'C'=2, and D'B'=3 ft. 

(3) Then the area of A'B'C'D'=D'B'x F'C, or 6 sq. ft. 



220 



FAIRCHILD'S SOLUTION BOOK. 



(4) We have 6 sq. ft. : 96 sq. ft :: 3 2 : DB 2 , from which DB = 

12 ft. 

(5) 6 sq. ft. : 96 sq. ft. :: 4 2 : AC 2 , from which AC = 16 ft. 

(6) CF = 8 ft., and DF= 6 ft. 

(7) .-. DC=VdF" 2 + FU 2 = 100 ft., the side of the rhombus. 

PROBLEM 436. 

The side of a rhombus is 10 and the shorter diagonal is 12: what is 
the area? 

Solution. 

(1) Let ABCD be the rhombus. 

(2) AB=10, DB = 12 and DF=6. 

(3) FCD is a right triangle and DC is the hypothenuse. 

(4) FC=DC 2 — DF 2 =8. 

(5) Area of ABCD = DBXFC, or 96. 

PROBLEM 437. 

How many acres in a field, two sides being- parallel, t) rd. and 14 rd., 
and the width 16 rd.? {Royer.) 

Solution. 

(1) Let ABCD represent the 

field. 

(2) AB = 14, DC=6 and CE = 

16 rd. 

(3) Area of ABCD = 

l(DC-r-AB) X CE = 160 

sq. rd., or 1 A. fig. 109. 

Rule. — Multiply half the sum of the parallel sides by the altitude. 
PROBLEM 438. 

Area of a trapezoid is 400, one of the parallel sides is 24, and the 
width 20: find the other side. 

Solution. 

(1) Let ABCD represent the trapezoid. 

(2) AB=24, and EC=20. 

(3) Then, 400-|-20=:20, and 20x2 = 40, the sum of the parallel 

sides. 

(4) .-. 40—24=16, or DC. 

Rule. — The sum of the parallel sides equals the area divided by 
the width, multiplied by 2. 




MENSURA TION 



221 



PROBLEM 439. 
The parallel sides of a trapezoid are 36 and 24, and the other two 
sides are each 10: find the width. 

Solution. 

(1) Let ABCD represent the trapezoid. 

(2) DC=24, AB=36, CB=AD=10. 

(3) AS = EB=(36— 24)^-2=6. 

(4) Now, CEB is a right triangle. 

(5) EC=VCB 2 — EB a =8. 

.*. 8 is the width of the trapezoid. 



PROBLEM 440. 
The sides, a=6, 6=4, c=5 and d=S y of a trapezium 
circle, being given, find the area of the trapezium. 

Solution. 

(1) Let ABCD be the trapezium. 

(2) AB = 6, BC=3, CD = 5 and DA 

=4. 

(3) Then by Bowser's Trigonom- 

etry, Art. 106, we have S = 
area and s=\(a-\-b-{-c-{-d). 



bed in a 



(4) S = V (s—a) (s—b) (s—c) {s—d) 
18.97. 




fk;. in 



PROBLEM 441 



What is the edge of the largest cube that can be cut from a hemi- 
sphere 20 in. in diameter? 

Solution. 

(1) Let ALB be the hemi- 

sphere, and DE the in- 
scribed cube. 

(2) ES = ;r, the e6ge of the 

cube. 



(3) SP=OP: 



+*• ° r 2 



(4) OF = 10 in., ESO is a 

right triangle. 

(5) 0? 2 = OP 2 +SP2. 

(x\* fxy 2x* 




FIG. 111. 



FAIRCHILD'S SOLUTION BOOK. 



(7) * 2 -H^ 2 = OE 2 , or 102. 

(8) ,r=8.1649 in. 

.-. 8.1649 in. is the edge of the cube. 

PROBLEM 442. 

A tin vessel having- a circular mouth 9 in. in diameter, a bottom 
4^ in. in diameter, and a depth of 10 in., is \ part full of water: what 
is the diameter of a ball which can be put in and just be covered by 
the water? 

Solution. 

(1) Let AVB represent a middle 

section of the vessel. 

(2) AB = 9 in:, UC=4.5 in, EF 

= 10 in. 

(3) The triangle BHC and BEV 

are similar. 

(4) BH : BE :: HC : EV, or 2\ 

: 4^ :: 10 : EV. 

(5) From this EV=20 , AV= 

VXE"2 + EV" 2 = 20.5. 

(6) 9 2 X.7854x' 2 3°=424.116cu.in., 

solid contents of the vessel. 

(7) (4|) 2 X .7854 X ^ =53.0145 

cu. in., solid contents of the 

T\\Tr FIG. 112. 

cone DVC. 

(8) QZ=radius of the ball that will just go into the cone. 

(9) |(9x20)=90 sq. in, area of the triangle AVB. 

(10) 9+(20.5x2)=50 in, the perimeter of AVB. 

(11) Since the radius QZ of the inscribed circle is found by 

dividing the area by half the perimeter, we have QZ= 
90-^25 = 3.6 in. 

(12) This is the radius of the inscribed sphere of the cone 

AVB, or 3.6x2 = 7.2 in., diameter of the ball that will 
just go into the mouth of the vessel. 

(13) (7.2) 3 X. 5236 = 195.4326528 cu. in., solid contents of the 

ball Q. 

(14) 424.116— 53.0145=371.1015 cu. in, solid contents of 

ADCB. 

(15) 424.116—195.4326258 cu.in. =228.6733472 cu. in, solid con- 

tents of whole cone — solid contents of a ball that just 
goes into it. 




MENSURATION. 



223 



(16) \ of 371.1015 = 92.7753 cu. in. = amount of water given in 

the vessel. 

(17) 92.7753+53.0145 = 145.7898 cu. in. = amount of water 

given in the vessel-f-the solid contents of the small 
cone DCV. 

(18) MN represents the surface of the water after the ball O 

is dropped in. 

(19) Now by similar solids, we have 228.6733472 : 145.7898 :: 

(7.2) 3 : (?) or GK, from which GK = 6.196+ in., the 
diameter of the ball O. 

NOTE. — If the ball O was 5 in. in diameter it would rest on the bot- 
tom of the vessel. 



PROBLEM 443. 

Find the diameter of the largest sphere that can be put in a hollow 
cone whose internal base is 2 ft. and altitude 3 ft. 

Solution. 

(1) Let MNV represent a middle section of the cone, and the 

inscribed sphere whose center is O. 

(2) Then MN = 2 ft., KV=3 ft., MV= V (3 2 -fl 2 ) =3.16227 ft. 

(3) Dividing twice the area by the perimeter of MNV gives 

radius OG; or OG =(3x2) -=-2+ (3.16227 X 2) =. "720705 ft. 

(4) Hence, the diameter is 1.44141 ft. 

PROBLEM 444. 

A conical wine-glass whose base diameter is 6 in. and altitude 5 
in., is filled with wine: if it be turned through an angle of 45°, how 
much wine will run out? 

Solution. 

(1) AB= diameter of the 

mouth, CF the depth 
and AMDN the surface 
of the wrne. 

(2) Draw CP perpendicular 

to AD; then the water 
remaining equals an ob- 
lique cone with altitude 
PC and base an ellipse 
whose axes are AD and 
NM. 

(3) Draw FO and EG parallel 

to KC, and through O 
pass a plane perpendic- 
ular to CF. 




FIG. 113. 



224 



FAIRCHILD'S SOLUTION BOOK. 



(4) The angle AEF=45°; then FE=AF 
CF— FE=2 in. 



3 in., and CE 



(5) From AEF and CEP, AE : CE : 

(6) From FCB and FGE , FC : FE : 

(7) AE=V(AF2— FE2)=3V2. 

(8) AG=AF+FG=4£. 

(9) From AGE and AOF, AG : AF 
(10) From FBC and CIK, CF : CK :: 



EF : CP=V2. 
FB : FG=f 



AE : AO 



¥V2. 



FB 



IK=¥ 



HK. 



(11) OH = HK- OK=|. 

(12) From geometry, OM=V(OHxIO)=-f. 

(13) AMDN = AOxMOX)r={f7rV2. 

(14) The cone ACD isequal toff7rV2x PC-h3=5.8905+ cu.in. 

(15) Cone ABC— ACD = 41.2335 cu. in., the amount of wine 

that will run out, 

NOTE. — This solution was prepared for the School Visitor by Prof. 
R. H. Sunkle. 



PROBLEM 445. 

Required the largest cube that can be placed in a conical cup 20 in. 
deep and 12 in. in diameter. 

Solution. 



(1) Let ACB represent a section 

of the conical cup and 
cube, where EF represents 
the diameter of the cup on 
which the lower side of the 
cube rests and also the 
diagonal of the bottom of 
the cube. 

(2) DC = 20 in., DB=5 in. 

(3) Let the edge of the cube=#. 

(4) Then DC : DB :: TC : TF, or 

20 : 6 :: 20— * : (120—6*) 
-r-20. 

(5) But the diagonal of the side 

of thecube is (120— 6*) 4-10 
—x^/2 f whence we find x= 
5.957+ in. 




FIG. 114. 



MENSURA TION. 



225 




FIG. 115. 



PROBLEM 446. 

Find the maximum cylinder which can be inscribed in a conical 
cup 1 ft. deep and 10 in. in diameter. 

Solutio?i. 

(1) Let ABC be the cone, and 

MN the inscribed cyl- 
inder. 

(2) Let BD=£, AD=tf, HD 

— x, FH=jj/, and the vol- 
ume of the cylinder=V. 

(3) We have v=iry*x. . . (1). 

(4) From the similar trian- 

gles ADBand AHF, we 
have AD : BD :: AH : 
FH, or a : b :: a — x : y. 

(5) .\y=— (a — x), which in 

(1) gives 
»=«£(«-, xYx . . (2). 

(6) Dropping constant factors, we have u=(a — x) 2 x=a 2 x — 

2ax 2 +x*. 

(7) .*. ~—a 2 — ±ax+3x 2 =0, or x*— $ax=— \a 2 . 

(8) .*. x—a, or \a. 

dn 2 d 2 it 

(9) -r^ — — ±a-\-6x; -j— 2 =2a, when x=a, .'. minimum. 

d 2 u 

(10) -j— 2 = — 2#, when x — \a, .'. maximum. 

(11) Hence, the altitude of the maximum cylinder is J of the 

cone. 

(12) The second value of x in (2) gives V = — -(a — £#)q= 

(13) Volume of the cone = ^7r#£ 2 . 

(14) .*. Volume of cylinder=| volume of cone. 

(15) y — ~{a — Jtf)=f£= radius of base of cylinder. 

(16) From the above the required result can easily be found. 



PROBLEM 447. 

From a cone, altitude 30 in. and radius of base 5 in., a 6-inch cyl- 
inder is cut as long as possible; from the top of the cone remaining- 



226 



FAIRCHILD'S SOLUTION BOOK. 




another cylinder is cut, of the same length as the former and the thick- 
est possible: how much of the volume of the original cone is cut away? 

Sohition. 

(1) Let ABC represent the cone, 

and SKDE and FETP the 
cylinders. 

(2) AO = 30in., OC=5in., LD = 

3 in. 

(3) COA and DLA are similar 

triangles. 

(4) We have CO: AO::DL:AL, 

or 5 : 30 :: 3 : AL. From 
which AL = 18 in. 

(5) LO = 30 in.— 18 in. = 12 in., 

the length of the cylinder 
SE, also the length of the 
cylinder FT. 

(6) ThetrianglesALDandTED 

are similar. 

(7) WehaveAL:LD::TE:ED, 

or ED=2 in. 

(8) LE=LD— ED=1 in., or ra- 

dius of small cylinder. 

(9) Volume of the large cylinder is (3 2 ttX 12) =108tt cu. in. 

(10) Volume of the small cylinder is (1 2 ttX 12) =12tt cu. in. 

(11) Volume of both cylinders=108ir+12ir = 120ir cu. in. 

(12) Volume of the cone ABC = 5 2 7rXl0=2507r cu. in. 

(13) Volume cut away = 2507r— 1207r=130ir, or 408.4 cu. in. 

PROBLEM 448. 

What are the two dimensions of a conical vessel that contains 13 
gallons of wine, the altitude being- 12 in., and the area of the top being - 
to the area of the bottom as 5 is to 3? 

Solution. 

(1) This problem is plainly solved by arithmetic. 

(2) The volume of the vessel is 231X13, or 3003 cu. in., which 

is the product of the sum of the three basal areas and \ 
of the altitude of the frustum. 

(3) Hence 3003-^4=750.75 sq. in., sum of the areas of lower, 

upper and mean bases. 

(4) Let these areas be as 3 : 5, and V~15 = 3.8729. 

(5) Then the lower base is 750.75 X (3-K1.8729) =189.695 sq.in. 

(6) The upper base is 750.75 X (5-^11.8729) =316.159 sq. in. 



FIG. 116. 



MENSURA TION. 



227 



(7) The lower diameter is V (189.695-i-.7854) =15.54 in. 

(8) The upper diameter is V (316.159-^.7854) =20.07 in. 



PROBLEM 449. 

How far must a fly walk on the shortest route from a lower to the 
opposite upper corner of a room 30 ft. long-, 25 ft. wide and 15 ft. high? 
Locate the point where it leaves the floor. 

Solution. 

(1) Let AE be the room. 

(2) AB = 30 ft., the length. 

(3) AG=25ft, the width. 

(4) AF = 15 ft., the height. 

(5) Let G be the corner at 

which the fly starts, 
and C the corner to 
which the fly travels. 

(6) Suppose the side 

AFCB to be lowered 
as a door upon its 
hinges until it forms 
a plane with the floor. 

(7) The shortest path the fly can travel is on the line GC. 

(8) GF'=25+15 = 40 ft., F'C'=30 ft., and the triangles GAP 

and GF'C are similar. 

(9) GC'=V"(W" 2 +FX'2)=:50 ft., the shortest path the fly 

can travel. 
(10) GF' : F'C :: GA : AP, or 40 : 30 :: 25 : AG, or 18| ft., the 
distance the fly leaves the floor from the corner A. 




fig. in 



PROBLEM 450. 

A cone 1 ft. high and 6 in. in diameter is perforated by a 2-inch 
hole, entering at the center of the base and passing through perpendic- 
ular to the base: find the part of the cone removed. 

Solution. 

(1) From Fig. 116, let ABC be the cone, and XP the 2-inch 

hole. 

(2) AO=12 in., and OC = 3 in. 

(3) Then we have the similar triangles ANT and AOC. 

(4) CO : OA :: TN : NA, or 3 : 12 :: 1 : NA, or 4 in. 

(5) NO = AO— AN=8 in., the height of the cylinder. 

(6) 8 xl 2 X7r=87rz= volume of the cylinder. 



228 FAIRCHILD'S SOLUTION BOOK. 

(7) Volume of the small cone APT is iX4Xl 2 X7r=~. 

o 

(8) Total volume removed is 8ir+- «-, or -=-. 

(9) Volume of the cone ABC is iXl2X3 2 X7r=36*\ 

4tt 

(10) The total volume removed is -~ -t-36tt, or -fa part of it. 

o 

PROBLEM 451. 

What is the distance from one lower corner to the opposite upper 
corner, of a box 12 ft. long-, 6 ft. wide and 4 ft. deep? 

Solution. 

(1) From Fig 117, let ABF be the box. 

(2) AB = 12 ft., BC=4 ft., and BD=6 ft. 

(3) ABD and ADE are right triangles, of which AD is the 

hypothenuse of the triangle ABD, and also the base of 
the triangle ADE. 

(4) AD 2 =AB2_{-BD 2 . 



(5) AE=VAD 2 +ED 2 = 14 ft., the diagonal of the box. 

PROBLEM 452. 

A vine passes around a pole, which is 52 ft. long, and 3 ft. in cir- 
cumference, once in every 4 ft. of its length: find the length of the 
vine. 

Solution. 

(1) One round of the vine forms the hypothenuse of a right- 

angled triangle, whose base is the circumference of 
cylinder, and perpendicular 4 ft. 

(2) Hence each round takes V(3 2 +4 2 )=5 ft. 

(3) Since there are 52-^-4=13, number of rounds, the vine 

must be 13X5=65 ft. long. 

NOTE. — Each round or spire is equivalent to the hypothenuse of a 
right triangle. This may be shown by covering a pencil or cylinder 
with paper and tracing the position of the vine upon it. Then unwrap 
the paper from the cylinder, and the right triangles will be seen. 

PROBLEM 453. 

In a cubical room a line drawn from an upper corner to the oppo- 
site lower corner is 24 ft.: find the size of the room. 

Solution. 

(1) From Fig. 117, let BF be the room, and AE the diagonal. 

(2) Let *=the edge. 



MENSURATION. 229 



(3) The n AD, the diagonal ofbase = V (AB 2 +DB 2 ), or 

V(* 2 +.* 2 ) = V2* 2 =W2. 

(4) EA 2 =AD 2 +ED\ or (*V2) 2 +* 2 =3* 2 . 

(5) 3* 2 =576, or * 2 =192. 

(6) x=sjV&, or 13.85+ ft. 

PROBLEM 454. 

If a cu. yd. of Klondike gold were formed into a bar \ in. square, 
without waste, what would be the length? 

Solution. 

(1) A cu. yd. contains 1728x27 cu. in. = 46656 cu. in. 

(2) The area of one end of bar is (^) 2 , or \ sq. in. 

(3) The length must be 46656+J = 186624 in. 

PROBLEM 455. 
A 16-inch tile will conduct as much water as how many 4-inch tiles? 

Solution. 



(1) Formula: Ratio of similar sides=V (ratio of areas). 

(2) The real quantity discharged is proportional to the areas 

of the ends of the tile, and since all circles are similar, 
we have from the above formula, 16 2 -^4 2 , or 16 times 
as much. 

PROBLEM 456. 

If cloth for a suit of clothes for a man weighing- 125 lb. costs $10, 
what will be the cost of enough of the same quality to make a suit for 
a man weighing 216 lbs., the men being similar in form and the suits 
similar in style? 

Solution. 

(1) Similar solids are to each other as the cubes, and the sur- 

faces as the squares of their like dimensions, or their 
homologous lines. 

(2) Therefore, we have (^I^T 2 : (^ISlS)"* :: $10 : $14.40, 

cost of the second man's suit. 

PROBLEM 457. 

The surface of a pyramid is 560 sq. in: what is the surface of 
another similar p3'ramid whose volume is 9 times as great? 

{Putnam Co. Test.) 
Solution. 

(1) From the above solution, (f/T) 2 : (f/9)* :: 560 : 2420+ 
sq. in. 



230 



FAIRCHIL&S SOLUTION BOOK. 




PROBLEM 458. 

The whole surface of a cylinder is 10 sq. yd.; the diameter equals 
the height: find the solidity. 

Solution. 

(1) Let AC be the cylinder, AB=^, 

and OE = CB=r.*. 

(2) Area of the circle or base AOB 

—\x 2 tt, and the area of both 
bases —\2x 2 tt. 

(3) ^7r=circumference of the base. 

(4) ^ 2 7r=convex surface. 

(5) 2(^ 2 7r)-|-^ 2 7r=whole surface. 

(6) .'. 2(^ 2 7r)+^ 2 7r=10 sq. yd. 

(7) * 2 =2.1220 sq. yd., and #=1.46+ 

yd.=AB = OE = CB, thediam- 
eter and height. 

(8) (1.46+) »x .7854= 1.665+ sq. 

yd., area of base. 

(9) 1.665+ X 1.46+ = 2.4236 cu. yd., solidity of the cylinder. 

PROBLEM 459. 

I have three farms containing 192, 216 and 168 A. respectively. I 
wish to divide these into fields of equal size: how many fields in all, if 
they are the smallest possible? 

Solution. 

(1) The G. C. D. of 216, 192 and 168 = 24, the number of acres 

in a field. 

(2) 168-^24=7, 192-^-24=8, and 216-^24=9 fields, respect- 

ively, making in all 24 fields. 

(3) The side of each field is V 24x160=56+ rd. 

PROBLEM 460. 

I place a bowl into the storm, 

To catch the drops of rain; 
A half a globe was just its form, 

Two feet across the same. 
• The storm was o'er, the tempest past, 

I to the bowl repaired; 
Six inches deep the water stood, 

It being measured fair. 
Suppose a cylinder, whose base 

Two feet across within, 
Had stood exactly in that place: 

What would the depth have been? 






MENSURA TION. 



231 



Solution 

(1) AB = 2 ft., the diameter of 

the globe. 

(2) SKMP = segment of the 

sphere, or the amount 
of water in the half globe. 

(3) PK=6 in., the depth of 

the water. 



(4) SK=VS0 2 — KO 2 

(5) Now as KO=6 in. 



SK= 




V12 2 — 6 2 =13.392+. 

(6) Volume of half the cylin- 

der, SY=|(SK 2 ttXPK) 
= 324tt. 

(7) Volume of the small sphere PK 

(8) .*. The volume of water in the vessel =3247r-j-367r=3607r. 

(9) AO = the radius of the cylinder given in that data of the 

problem, then SK 2 7r=1447r, the area of the base of the 
cylinder, whose diameter is the same as the diameter of 
the half globe. 
(10) 3247r-i-1447r=2| in., the depth that the water would have 
been in the cylinder. 

PROBLEM 461. 

If a hole 12 in. in diameter be made through the center of a sphere 
whose diameter is 20 in., how many cu. in. of the sphere will be con- 
sumed? 

Solution. 

(1) Let EDMS be the 

sphere, and EFMSbe 
the cylinder made by 
the auger. 

(2) Then the volume bored 

out consists of a cyl- 
inder and two spher- 
ical sections, of which 
E D F, SMP.and 
EFMS are vertical 
sections. 

(3) SO = 12 in., the radius 

of the sphere. 

(4) SK=6 in., the radius FIG ,->„ 




232 



FAIRCHIL&S SOLUTION BOOK. 



of the auger hole. 

(5) PK=the diameter of the small sphere. 

(6) KO=V(S0 2 — SK 2 ), or 8 in. 

(7) PK=PO— KO=2 in. 

(8) The volume of half the cylinder SMYZ=|(SK 2 7rX KP) 

= 36tt. 

(9) The volume of the small sphere PK=PK 3 X-=^". 

o 

(10) (36ir+y\ X 2=^1^=234.5728 cu. in.=volume of both 

segments = EDX+SMP. 

(11) Volume of the cylinder SF= (EX 2 XttX KX) =1809.562 

cu. in. 

(12) The amount consumed is 234.572+1809.562=2044.134+ 

cu. in. 

Rule for finding a Spherical Segment. — A spherical segment with 
one base is eqtiivalent to half a cylinder having the same base and 
altitude, phis a sphere whose diameter is the altitude of the seg- 
ment. 

PROBLEM 462. 

By boring- through the center, what is the diameter of an aug-er 
that will bore away one-half of a ball 6 inches in diameter? 

Solution. 

(1) Let O be the center of 

the ball. 

(2) OB=r=3, OF=CD= 

R, the radius of the 
auger bit, and CB=^. 

(3) ThenCT) 2 =R 2 = 

(2r — x)x, and CO = 
r — x. 

(4) The volume removed 

equals a cylinder, ra- 
dius R, and height 2(r — x), together with two segments 
of the ball of height x, and radius R. 

(5) The volume of the cylinder is 2(2r — x) XTr(r—x). 

(6) The volume of the two segments is ^ttx 2 (Qr — 2x). 

(7) Adding and equating to half the volume of ball fr 2 *-, we 

have 4* 3 — 12^ 2 +12r 2 ^ 2 — 4r 3 =— 2r 3 . . . (1). 

(8) (l)-^4 = ;t: 3 — 3r-* 2 +3r 2 ;t 2 — r 3 = — \r* . . . (2). 




FIG. 121. 



MENSURA TION. 233 

(9) &JZ)=x— r=rf% . . . (3)._ 

(10) .-. R* = ( 2r— x)x=r *(l — jf2) . . . (4). 

(11) 2R=2rV (1— iV2) =6V (1— if 2), diameter of auger bit, 

or 3.6496 in. 

PROBLEM 463. 
A blacksmith had an iron ball, 
Which he did a bullet call; 
He found out, by some nice trick, 
It was exactly one foot thick; 
And silvered o'er with plate, 
And which he proved by scales and weight; 
The silver's weight he did pronounce, 
Was just exactly half an ounce; 
He took the ball and made it hot, 
And forged there out an iron rod; 
The rod was just three inches thick, 
Round and long, just like a stick; 
Now how much silver will it take 
To plate the rod that he did make, 
And put it on exact as thick 
As that which on the ball did stick? 
Solution. 

(1) The volume of the ball is 12 8 X^, or 288tt cu. in. 

(2) This is also the volume of the cylindrical rod. 

(3) The area of the end of the rod is (§ ) 2 7r=2^7r. 

(4) The length of the rod must be 288iH-2£*r, or 128 in. 

(5) 37rXl28 = 3847r sq. in., the convex surface of the rod. 

(6) The basal surface is 2^7rX2 = 4^7r sq. in. 

(7) 384tt +4-^=388^ sq. in., the surface of the rod. 

(8) 12 2 X7r=1447T sq. in., the surface of the ball. 

(9) We have by proportion, 144tt : 388|tt :: | oz. : (Iff J- oz.). 

.'. It will take If 2 J oz. to plate the rod forged from the 
iron ball. 

PROBLEM 464. 

A leaden spherical shell, hollow part 6 in. in diameter and thick- 
ness of solid part 3 in., is converted into a three-inch pipe, outer meas- 
ure 5 in.: find length of pipe. 

Solution. 

(1) Since the solid part is 3 in. thick, the shell is 6+3x2 = 

12 in. 

(2) Suppose this sphere to be solid; then its volume would 

be 12 3 X1tt, or 288tt cu. in. 



234 FAIR CHILD'S SOLUTION BOOK. 

(3) The diameter of the hollow part is 6 in.; then the vol- 

ume of the hollow part is 6 3 X^tt, or 36tt cu. in. 

(4) 288tt— 367r=2527r, cu. in. of lead in the shell. 

(5) The area of the lead pipe at one end is (-f) 2 7r — (-|) 2 7r= 

J ¥ 6 -7r, or 4:7r sq. in, 

(6) .-. The length of the pipe is 252^^-4^=63 in. 

NoTK. — It would be well for the student to learn to use the charac- 
ter 7T instead of 3.141,592,653,989,793,238,462,643,383,279,502,884,197,169,- 
399,375,105 . . , for much labor can be saved, as in the above solution. 
For example, 7r-s-7r=l. F or practical purposes it is sufficiently accurate 
to call 7T=3.1416, or 3.14159. Now, 3.1416-^3.1416=1; 20^-^2tt=10. 

104. The Origin of ir. — At least from the time of Archi- 
medes, 7r has stood for the number expressing how many 
times the diameter the circumference is. It is the initial let- 
ter of the Greek word -n-epicfrepa, meaning periphery. If the 
diameter is taken as a unit, then it stands for the periphery, 
or circumference. 

Prof. W. W. Beman, of the University of Michigan, says that t 
was first used to represent the number 3.141592 in Jones' Synopsis Pal- 
mariorum, London, 1706. This fact seems to have been overlooked by 
most writers on mathematics, as we find the statemeut that Fouler was 
apparently the first to use tt, Encyclopedia Britannica, ninth edition, 
(A. M. M.) 

Archimedes, 250 B. C, the greatest mathematician of ancient 
times, proved that the value of it lies between 3f and 3£t> that is, be- 
tween 3.1429 and 3.1408. Ptolemy, 150 B. C, used the value 3.1417. 
Metrus, of Holland, used the fraction fff , which is correct to six places 
of decimals. Lambert, 1750 A. D., proved tt incommensurable. 

Many hare-brained visionaries still attempt to square the circle, 
and will continue to chase after this will-o'-the-wisp. Squaring the 
circle, that is, finding a square equal in area to a circle of given radius, 
has called out an untold amount of work, but all attempts have ended 
in total failure. The oldest known mathematical work, the Rhind 
Papyrus (c. 2,000 B. C.) contains the problem in the well known form, 
to transform a circle into a square of equal area. The writer of the 
papyrus, Ahmes, lays down the following rule: Cut off \ of a diameter 
and construct a square on the remainder; this has the same area as the 
circle. The value of ir thus obtained is (-^) 2 =3.16, not very inexact. 
Still farther from the correct value is that of 7r=3, which is found in 
the Bible. I. Kings 7: 23 and II. Chron. 4: 2. 

We see all this expended mental energy has been fruitless, but 
nevertheless has worked for advancement in the manifold realm of 
mathematics. 



MENSURATION. 235 

PROBLEM 465. 

Four ladies own a ball of thread 3 in. in diameter: what portion of 
the diameter must each wind off in order to have equal shares of the 
thread? 

Solution. 

(1) Volume of the ball is 3 3 X -^-=14.1372 cu. in. 

(2) \ of this is 3.5343 cu. in., the share of each lady. 

(3) The last lady has a ball containing 3.5343 cu. in., whose 

diameter is the cube root of (3.5343-K5236) =1.89 in. 

(4) After the first and second have wound their shares off, 

the last two have a ball containing (3.5343X2) =7.0686 
cu. in. 

(5) Its diameter is ^(7.0686-^.5236) =2.38+ in. 

(6) Then 2.38— 1.89 = .49 in., third lady's share. 

(7) After the first lady has wound her share off, the three re- 

maining ladies will have a ball whose volume is 
( 3.5343x3) =10.602 9 cu. in., and whose diameter is 
^(10.6029-^.6236) =2.72 in. 

(8) Then 2.72— 2.38=.34 in., the second lady's share. 

(9) 3— 2.72=.28 in., first lady's share. 

.'. The first winds off .28, the second .34, the third .49 and 
the fourth 1.89 in. 

PROBLEM 466. S a *V 3 *V~ ^ 

John has a ball of yarn 10 in. in diameter and a wooden ball 6 in. 
in diameter: how thick will be the yarn when wound on the wooden 
ball? 

Solution. 

(1) The volume of the ball of yarn is 10 3 X \^—~^ 3 -^ cu. in. 

(2) The volume of the wooden ball is 6 3 X^7r=367r cu. in. 

(3) The combined volume of the balls is 1 V J ^+367r=202f7r 

cu. in. 

(4) The diameter is V202f7r-^7r= 10.6736 in. 

(5) The radius of the ball is \ of 10.6736 = 5.3368 in. 

(6) The radius of the wooden ball is \ of 6 = 3 in. 

(7) .'. The thickness of the yarn is 5.3368—3 = 2.3368 in. 

PROBLEM 467. 

Find the volume of the largest square pyramid that can be cut from 
a cone 9 ft. in diameter and 20 ft. high? 

Solution. 
(1) The diameter of the cone is the diagonal of the largest 



236 FAIRCHILD'S SOLUTION BOOK. 

inscribed square, or the base of the required pyramid. 

(2) Hence, the diagonal 9 squared, divided by 2, gives 40.5 

sq. ft., the area of the base of the pyramid. 

(3) 40.5 multiplied by J of the altitude 20, gives 270 cu. ft., 

the volume required. 

PROBLEM 468. 

A cylindrical vessel 1 ft. deep and 8 in. in diameter was \\ full of 
water; after a ball was dropped into the vessel it was full: find the 
diameter of the ball. 

Solution. 

(1) The area of the base is 4 2 X7r, or Voir. 

(2) The volume of the cylinder is 167rXl2=1927r cu. in. 

(3) Now as the water occupied ^f of the volume, the volume 

of the ball must have been T 3 g of 1927r, or 36tt. 

(4) The cube root of (dQ-n-^l-v) =6 in., the diameter of the ball. 

PROBLEM 469. 

A circular table 6 ft. in diameter weighs 40 lb. and rests on four 
legs in its circumference, forming a square: find the least vertical 
pressure that must be applied to overturn it. 

Solution. 

(1) From Fig. 100, let ABCD be the circular table. 

(2) Let A, B, C, D, be the places where the legs are attached. 

(3) K is the point of application of pressure. 

(4) OL=3V2-^2=2.1213 ft. 

(5) LK = 3— 2.1213 = .8787 ft. 

(6) Let O be the center of gravity of the table; then, by a 

principle in Mechanics, all its weight may be consid- 
ered to be at O. 

(7) Then, L is the center of moments, and by the principle of 

moments we have, (Pressure required) = — p^ — == 

96.5+ lb. 

Note. — From the data given, the legs are supposed to be of equal 
length, and also to be without weight. 

PROBLEM 470. 

It is desired that three men carry each ^ of a 12 ft. log of uniform 
size and density. Where must the hand-stick be placed so that the one 
at the end of the log and the others at the ends of the stick, shall each 
carry equal weight? (Hancock Co. Test.) 



MENSURA TION. 



237 



Solution. 

(1) When a body is supported at two points, the weight sus- 

tained at the two points respectively is inversely as 
their distance from the center of gravity, which in the 
log corresponds with its center, or 6 ft. from each end. 

(2) The man at the end of the log is 6 ft. from its center. 

(3) Now, that the two men at the hand-stick may carry f of 

the log, or twice as much as the one at the end of the 
log, the hand-stick must be placed \ as far from the 
center of the stick, or 4 of 6 = 3 ft. 



.•. The hand-stick must be placed 3 
the log. 



ft. from the end of 



NOTE. — This seems to be more of a mechanical than an arithmet- 
ical problem, for the above is solved by a well known principle of 
mechanics. 

PROBLEM 471. 

A string- is wound spirally 100 times around a cone 100 ft. in diam- 
eter at the base: through what distance will a duck swim in unwinding 
the string, keeping it taut at all times, the cone standing on its base 
and at right angles to the surface of the water? 

Solution. 



(1) Let/; = height, r= 
radius of base, /= 
V (// 2 +r 2 )= slant 
height, n = num- 
ber of spirals. 

(2) Let D, E be the 
two consecutive 
poin t s in the 
duck's path. 

(3) K,L = correspond- 
i n g consecutive 
points of tangency 
of string to cone. 

(4) F, G = c o r r e s- 
ponding consecu- 
tive points in base 
of cone. 

(5) Let CD=^, FK= 
x. 

(6) DE=^, ML=^. 




FIG. 122. 



238 FAIRCHIL&S SOLUTION BOOK. 

(7) ^ 2 = DN 2 +NE 2 . . . (1). 

(8) From the similar triangles GOF and DFE, r : GF=FD 

: DN. 

(9) .-. DN=(FD.GF)-^ . . . (2). 

(10) From the triangles LMK and KFD, dx : MK=^ : DF. 

(11) FD=^.MK^^ . . . (3). 

(12) MK : (/— *) = GF : /. 

(13) MK=[(/-j).GF]^/ ... (4). 

(14) But CF : x=2irrn : I. 

(15) .'. GF : dx :: ^rn : l. 

(16) GY—^rn.dx)^ ... (5). 

(17) (5) in (4) gives, MK=[2tt^(/— x)dx]-±l* ... (6). 

(18) (6) in (3) gives, ¥T>=[2irrnx{l— ^)]-=-/ 2 . . . (7). 

(19) (7) and (5) in (2) give, PN=[4tt W(/— x)xdx]+rl* (8). 

(20) The increment of FD is, (GH + NE) =FG+NE. 

(21) .-.From (7), <tf(FD)=[2 w /w(/— 2*)^]^/ 2 = (FG+NE) 
. . . (9). 

(22) (5) in (9) gives, NE=[—47rmxdx]-+l* . . . (10). 

(23) (8) and (10) in (1) gives, ds-jf-^\+-±-(l-xYdx. 

XirVH CI I 7r 2 ^ 2 

(24) :, s = yPJ Jl+^L(/_^)2^ =(4 r/37r;0 + 

(2r/3)[7r;2— (2/7r«)]Vl+^ 2 ^ 2 +2Hog(7r^+Vl+7r 2 ?^). 

(25) In the problem, r=l, ;*=100. 

(26) .'. J=(l/757r)+|[1007r— (l/507r)]Vl + 100007r 2 + 

2 log(1007r+VH-100007r 2 ). 
.-. 5=68948.771 ft. 



X. TETRAHEDRON. 

105. A Tetrahedron is a solid having four equal faces, each 
of which is an equilateral triangle. 

The area of one face is found by squaring its edge and 
multiplying by V3=1.73205. If the edge is 1, then the sur- 
face of the tetrahedron is l 2 X V^X 4=6.928. 



MENSURA TION. 



239 



Its volume is found by 
multiplying ^ of the cube 
of an edge by v"2=1.4142. 
Hence, the edge is 1; we have 
13 Xt2><V2 = .11785, the vol- 
ume. 

Now, let us find the alti- 
tude D G. AB=1, AF=i 
DF= V(l 2 -(i) 2 )=lV3 = 
\ of 1.732, or .866, which 
equals FC, also FG is J of 
FC, or iV 3 = .28866. DG= 
DF 2 — FG 2 

V 



i 

mm 



or 



FIG. 123. 



.81649. 



[(iV3) 2 -(W3) 2 ]=v(f- T V) = Vf=W6 : 

The center of the inscribed sphere is also the center of the 
tetrahedron, which is in DG at O, J of DG from G is GO, or J 
of ^^y 6=^ V6=. 20412, the radius of the inscribed sphere. 



CHAPTER XIII. 



PROBLEMS. 



1. If .63 gallons of wine cost $1.47, what will -£$ of a gallon 
cost? Ans. $1.05 

2. James at college wrote to his parents that he had paid 
•§■ of f of his debts, and still owed $44: what were his debts? 

$128. 

3. If 4 pounds of cheese can be bought with 60 ct. when 
milk is 10 ct. a quart, how many pounds should be bought 
with 36 ct. when milk is 8 ct. a quart? 3 lb. 

4. Bought 5^ yd. of cloth at $4^ a yard, and paid for it 
with wheat at $l-f- per bushel: how many bushels were re- 
quired? 14 bu. 

5. A and B can do a piece of work in 12 days, and A can 
do f as much as B: in what time can each do it? 

A 30 days, B 20 days. 

6. If i of 6 be 3, what will T \ of 40 be? 54. 

7. A can build a wall in 18f days, B in 3LJ- days: how long 
will it take both together? 11|~| days. 

8. A man after doing f of a piece of work in 30 days, en- 
gages an assistant and both together complete it in 6 days: 
in what time could the assistant do it alone? 21|- days. 

9. | of (2H-»)+i of f =what? 2fJ. 

10. When flour is $6 a barrel, the 10-cent loaf weighs 30 oz.: 
what is the price of flour when it weighs 24 oz.? $7.50 a bbl. 

11. If f of a yard of ribbon costs $f, how many yards can 
be bought for $|? f of a yd. 

12. If a street vender buy 5 bu. of chestnuts for $18.50, and 
sells them at 15 ct. per liquid quart, how much does he gain? 

$9.43. 
(240) 



PROBLEMS. 241 

13. A merchant owns ^ of a store, and sells f of his share 
for $2100: what is the value of the part he still owns? $2800. 

14. A man owning -J of a mill, sold f of his share for $12000: 
what was the value of the mill? $45000. 

15. A, B and C do a piece of work for $79.35: on the 
supposition that A and B do f of the work, A and C T 9 ¥ , and 
B and C \%\ what should each receive? 

A $27.60, B $17.23 and C $34.50. 

16. Corda, being asked her age, replied: "My age! If it is 
multiplied by 3, and -| of the product tripled be, the square 
root of | of that is 4: now, tell my age and never ask me 
more." 28 yr. 

17. Four times £ of a number is 12 more than twice the 
number: what is the number? 18. 

18. I had | of my money stolen; I then earn $60, and 
spend \ of all I now have; my uncle then gives me $10, and 
after losing \ of what I have I find I have half as much money 
as I had at first: what had I at first? $75. 

19. Divide the L. C. M. of J, f , |, | and -f by the G. C. D. of 
83J and 268|: what is the quotient? 28$. 

20. 6+4x3— 5x2+6+2 x 8-^-4-9= what? 9. 

21. f of the square of J of a number is 12 more than -} of 
the square of half the number: what is the number? 8. 

22. Find the number, f of whose cube is 10 more than the 
cube of its J. 3. 

23. Squaring twice a number gives 18 more than twice its 
square: find it. 3. 

24. Evert went -f of a journey on a bicycle, f of the remain- 
der by coach, and walked the last 3 miles: how long was the 
journey? 42 mi. 

25. What is the number, to which if 15 be added and the 
sum multiplied by 9, and 11 taken from the product, the re- 
mainder will be 340? 24. 

26. If a hen and a half could lay an egg and a half in a day 
and a half, how many eggs would 6 hens lay in 7 days? 

28 eggs. 



242 FAIRCHILD'S SOLUTION BOOK. 

27. If 6 cats catch 6 mice in 6 minutes, at the same rate 
how many cats will it take to catch 100 mice in 100 minutes? 

6 cats. 

28. If an article had cost 10% less, my rate of gain would 
have been 15% more: what was my rate of gain? 15%. 

29. What must be the asking price of cloth, costing $3.29 
per yard, that I may deduct 12-J-% and still gain 12|%? 

$4.23. 

30. Cloth marked at 30% above cost price, having deprec- 
iaied in value, was sold at 28% off the marked price: what 
% did I gain or lose? I lost 6f%. 

31. John sold an article at 25% gain; had it cost him $1.25 
more, he would have lost 30%: what was the cost? $1.59 T 1 T . 

32. Had an article cost me 15% more, the same selling price 
would have given a rate of gain 12% less: find my rate of loss? 

8%. 

33. My merchant bought tea at 25% below wholesale cost, 
receiving a discount of 5% for cash payment, and sold at 12-|% 
above wholesale price: what per cent did he gain? 57^J%. 

34. A merchant bought a piece of goods, which shrank ^% 
both in length and width; he sold 7 yd. for what 6 yd. cost 
him: what was the loss %? 14^|%. 

35. A's farm is worth 25% more than B's, and B's is worth 
10% less than C's; if A should trade 75% of his farm for 80% 
of C's, he would lose $87.50: how much is B's farm worth? 

$1800. 

36. If the cost of an article had been 8% less, the gain 
would have been 10% more: find the % gain. 15%. 

37. If the cost had been 16f % more, the gain would have 
been 18f % less: what was the % of gain? 31^%. 

38. Bought 1000 bu. of corn at 16 ct. a bushel; if the loss 
by shrinkage and waste is 5%, for what must I sell per bushel 
to gain 12|% on my investment? $.18947. 

39. A mill is worth 5% less than a store, and the store 20% 
more than a house; the owner of the house has traded it 
for 75% of the mill, thereby losing $145: what was the value 



PROBLEMS. 243 

of the store? Of the mill? Of the house? 

Store $1200, mill $1140, house $1000. 

40. I sell two horses for $430, gaining 20% on one and los- 
ing 20% on the other; I lose $20 by the transaction: find the 
cost of each. The first horse cost $175, the second $275. 

41. I bought a certain number of hats at $2.50 each; I sold 
£ of them at 25% profit, and on the sale of the remainder I 
lost $15; my total gain was equal to 5%: find the number of 
hats. 120. 

42. Jesse sold two horses for the same sum; on the one he 
gained 20%, and on the other he lost 20% ; his whole loss was 
$25: what did each horse cost him? 

The first cost $250, the second $375. 

43. I sold an article at 20% gain; had it cost me $300 more, 
I would have lost 20%: find the cost. $600. 

44. I gained 120% by selling rice at 8 ct. per pound: what 
did the rice cost per pound? 3 T 7 r ct. 

45. If \ an article be sold for the cost of \ of it, what is the 
rate of loss? 33£%. 

46. A merchant sold part of his goods at a profit of 20%, 
and the remainder at a loss of 11%; his goods cost $1000 and 
his gain was $100: how much was sold at a profit? $677.41|f # 

47. At what price must an article which cost 30 ct. be 
marked, to allow a discount of 12-^% and yield a net profit of 
16|%? 40 ct. 

48. Mr. Marble bought a lot of cassimeres at $3.75, and 
marked them "njio^" his key being "John Marble": what was 
his gain % at the marked price? 68%. 

49. Mr. Smith's key for marking goods was "Republican:" 
if he marked some silks "e.bn" and gained at that rate ll-J-% 
on the cost, what was the cost per yard? $2.25. 

50. Prof. Zerr loses 16% by selling his library for $960 less 
than it cost: what must he have received for it if he had 
gained 16%? $6960. 

51. I remitted to my agent $508.95 to invest in potatoes, 
after deducting all expenses. He paid $46 for barrels, $38.60 
for drayage, and charged 2|% commission for buying: how 



244 FAIRCHILD'S SOLUTION BOOK. 

many barrels of potatoes did he buy at $2.25 per barrel? 

184. 

52. The Continental insured f of A's machine shop at 3%; 
the Phoenix relieves them of \ the risk at 2%, and the Teu- 
tonia of \ of it at 1|%. By fire the Phoenix lost $3250 more 
than the Teutonia: what was A's rate of loss on value of shop? 

35i%. 

53. A conical cup 6 in. deep and top 5 in. in diameter, is 
full of water; a ball is then dropped into it and is just im- 
mersed: how much water overflows? 19.30 cu. in. 

54. A round table 10 ft. in diameter has four drop leaves of 
equal size, and when the leaves are down, the table is a 
square: find area of one leaf. 7.133 sq. ft. 

55. A cylinder 60 ft. high and 4 ft. in circumference has a 
rope passing around it spirally, each round rising 3 ft.: how 
long is the rope? 100 ft. 

56. Find the diagonal of a rectangular field containing 13-J- 
acres, whose length is to its breadth as 12 to 5. 

Length 72 rd., breadth 30 rd., and diagonal 78 rd. 

57. A 12-inch tile will conduct as much water as how many 
3-inch tiles? 16. 

58. What is the convex surface and whole surface of a right 

cone whose slant height is 66 ft. 8 in., radius of the base 4 ft. 

2 in.? 

Convex surface 870.7 sq. ft., whole surface 924.61 sq. ft. 

59. In a right-angled triangle, having the side of the in- 
scribed square 36 and the radius of the inscribed circle 20, to 
determine the triangle. Base 84, perpendicular 63. 

60. The altitude of a triangle is 39 and base 21: required 
the side of the largest inscribed square. 13|f . 

61. What must be the diameter of a round log 12 ft. long 
which, when squared, will contain 54 cu. ft.? \\ ft. 

62. The area of a cube is 72 sq. in.: find its diagonal. 6 in. 

63. The surface of a cube is 200 sq. in.: find the surface of 
its circumscribed sphere. 314.15926 sq. in. 

64. The surface of a sphere that can just be enclosed in a 
cubical box is 400?r sq. in.: find inner surface of the box. 

2400 sq. in. 



PROBLEMS. 245 

65. A ladder 20 ft. long stands 12 ft. from the base of a 
building, and the top rests against the wall 4 ft. below the 
eaves: find the height of eaves. 8 ft. 

66. How many posts 7 ft. apart will it require to fence a 
rectangular lot containing 70756 sq. ft., the length of the lot 
being 4 times its breadth? 190. 

67. A circular park is crossed by a straight path cutting off 
\ of the circumference; the part cut off contains 10 acres: 
find the diameter of the park. 150 rd., nearly. 

68. If a post 4 ft. high casts a shadow 13 ft. in length, what 
must be the height of a post that will cast a shadow 125 ft. in 
length? 38 jk ft. 

69. The, perimeter of a right-angled triangle is 120 rd., and 
the radius of the inscribed circle is 10 rd.: find the sides. 

30, 40 and 50 rd. 

70. A lead cone is 4 in. in diameter and 1 in. high: find the 
diemeter of a bullet that can be made from it? 2 in. 

71. The diameter of a spherical balloon is 25 ft.: how many 
square yards of silk were required to make it, and how many 
cubic feet of gas will be required to fill it? {Putnam Co.) 

218i sq. yd. of silk, and 8181.2 cu. ft. of gas. 

72. If two boxes are similar, and one, with dimensions 2, 4 
and 5 ft. respectively, holds eight times as much as the sec- 
ond, what are the dimensions of the second? 

1, 2 and 2| ft. 

73. A cylindrical cistern is 10 ft. in diameter and 20 ft. 
deep: give contents in barrels. 384.44 bbl. 

74. Two circles have the same center. A chord, length 
20 ft., is drawn across the large circle, and it will just touch 
the circumference of small circle. Required the area of the 
ring. 314.16 sq. ft. 

75. Upon the diagonal of a rectangle 24 ft. by 110 ft., a tri- 
angle equivalent to the rectangle is constructed: find its alti- 
tude. 18 T 6 3 ft. 

76. The ends of a cord 100 ft. long are fastened to stakes 
placed 80 ft. apart on level ground; a ring to which a sheep 
is tied plays freely on the cord: find the area of the ellipse 
over which the sheep can graze. 1500tt sq. ft. 



246 FAIRCHILD'S SOLUTION BOOK. 

77. Three men, A, B and C, residing at the several vertices 
of a triangle, the sides of which are 65, 70 and 75 chains re- 
spectively, agree to build a school house the center of which 
shall be equally distant from the residences: what is the dis- 
tance from each residence to the school house? 40.625 ch. 

78. A tub of butter weighs 16 lb. on one end of a pair of bal- 
ances, and 25 lb. on the other: find the correct weight? 

20 lb. 

79. An equilateral triangle has an area of one acre, and is 
to be enclosed by a fence: what is the length of each side? 

19.2 rd. 

80. How far will the head of a nail in the tire of a wheel 
move in driving 78.54 rd.? 100 rd. 

81. What is the greatest amount of land that ?r rd. of rope 
will enclose? \tt. 

82. A tree 120 ft. high stands on the bank of a river 100 ft. 
wide: where must the tree break off so that it may remain 
connected at the point of breaking and its top just reach the 
opposite shore? 18 J- ft. 

83. The length and breadth of a school room are as 8 to 5/ 
but if it were 2 ft. longer and 2 ft. wider, the area of the floor 
would be 1600 sq. ft.: find dimensions. 30 and 48 ft. 

84. A hawk and a dove are in the air, the hawk 150 ft. above 
the dove: if the dove flies straight forward in a horizontal line, 
and the hawk flies \\ times as fast directly towards the dove, 
how far will the hawk fly before it catches the dove? 

The hawk flies 416f ft., the dove 333J ft. 

85. A wealthy man two daughters had, 

And both were very fair ; 
To each he gave a tract of land, 

One round, the other square. 
At twenty shillings an acre just 

EJach piece its value had; 
The shillings that did encompass each 

For it exactly paid. 
If 'cross a shilling be an inch, 

Which, had the greater fortune, 
She that had the round or square? 



PROBLEMS. 247 

86. If the sun crosses the equator at 5:20 a. m., Central 
Standard time, in what longitude does it cross if the sun is 10 
min. slow? 7\° east of Greenwich. 

87. If the sun crosses the equator at 7:10 a. m., Central 
Standard time, in what longitude does it cross if the sun is 10 
min. slow? 20° west of Greenwich. 

88. A mountain known to be 4 mi. high was seen from a 
ship 178 mi. distant: from this data calculate the diameter of 
the earth. 7917 mi. 

89. A sells goods which cost him $160, at a certain rate of 
gain; B sells to C at the same rate of gain; C pays $360 for 
the goods: find the rate of gain for A and B. 50%. 

90. In a circle 2 ft. in diameter, a five-pointed star is in- 
scribed. The length of one side of the star is 8 in.: what area 
is covered by the star? 136.1 sq. in. 

91. If a sphere, the diameter of which is 4 in., is depressed 
in a conical glass £ full of water, the diameter is 5 in., and 
altitude 6 in.: how much of the vertical axis of the sphere is 
immersed? .544 in., nearly. 

92. The area of a field is 30 times its diagonal, and its sides 
are in the ratio of 3 to 4: find diagonal and sides. 

Diagonal 62£, sides 37-^ and 50. 

93. Two trees 48 ft. and 60 ft. high, respectively, are stand- 
ing 50 ft. apart: at what point between them must a boy stand 
that he may just fly his kite to the top of either tree? 

12.04 ft. 

94. In a store the "key" by which goods are marked was 
"panegyrist," the letters denoting the cost being above a line 
and those denoting the selling price below. Using this key as a 
guide, supply the blanks in the following, the gain in each 

being 20%: -.-^T. H &Z 

an - m' peg' 

95. If the major axis of an ellipse is 200 ft., and the minor 
axis 120 ft., what is the distance between the foci? 160 ft. 

96. A man sold two horses at equal prices; on one he 
gained 25%, and on the other he lost 25%; he lost by the 
transactions $20: find the cost of each horse. 

$120 and $200. 



248 FAIRCHILD'S SOLUTION BOOK. 

97. A street passes through a semicircular arch under a 
railroad. If it would take 777| perches of masonry to fill up 
the entire archway, it being forty feet long, how wide is the 
street? 35 ft. 

98. What annual rate of interest on the net sales is equiva- 
lent to a discount of 5% on the gross, if the term of credit is 
60 days? 31^%- 

99. What % in advance of the cost must a merchant mark 
his goods, so that after allowing 4% of his sales for bad debts, 
an average credit of 3 months, 10% of the cost of his goods 
for expenses, he ma} 7 make a clear gain of 14% of the first 
cost of the goods, money being worth 8% ? 31f %. 

100. The difference between the simple interest and the true 
discount for 4 years is $llf , and for 8 years is $40: what is the 
rate % and sum of money? 5% and $350. 

101. If the remainder is 17 and the dividend 4352, what is 
the divisor and the quotient, if the remainder is the G. C. D. 
of the divisor and quotient? 51 and 85. 

102. How much more water will a tile 6 in. in diameter dis- 
charge per hour than one 4 in. in diameter? 2-J- times. 

103. A merchant marks his goods at 40% in advance of 
cost, and selling uses a pound weight \ oz. too light; if he 
throws off 10% of his marked price, find the gain per cent. 

26«f%. 

104. Ten men hire a coach; by getting 4 more passengers the 
expense to each is diminished %\\\ what do they pay for the 
coach? $42. 

105. A well is 150 ft. deep and 110 ft. in diameter: if it is 
one-third ; full of water, how much work is required to empty 
it? 30679598.42 ft. lb. 

106. In tne midst of a meadow, well covered with grass, 

Just an acre was needed to tether an ass; 
How long was a line that reaching- all 'round, 
Restricted its grazing - to an acre of ground? 

39.2507 yd. 

107. If the area of an equilateral triangle is 300, what is the 
side? 26.32. 



PROBLEMS. 249 

108. A coin is held 5 ft. from a wall and parallel to it; a 
luminous point 15 in. from the coin throws a shadow of it 
upon the wall: how does the size of the shadow compare with 
that of the coin? The shadow is 25 times the coin. 

109. The hypothenuse of a right-angled triangular piece of 
land containing 15 acres is 100 rd.: what is the length of the 
sides? 60 and 80 rd. 

110. A boy is running on a horizontal plane directly towards 
the foot of a tree 50 ft. in height: when he is 100 ft. from the 
foot of the tree, how much faster is he approaching it than 
the top? 1.118 times. 

111. Find the cost of 20 planks, each 16 ft. long, 10 in. wide, 
and 2\ in. thick, at $15 per M. $10. 

112. I have a circular piece of ground 160 ft. in diameter, 
from which I take enough earth to fill a hollow cone, whose 
base is 16 ft. in diameter and whose depth is 30 ft.: how much 
is the lot lowered? -^ ft. 

113. How many quart, pint and half-pint bottles of each an 
equal number, can be filled from a vessel containing 8 gal. 
3 qt.? 20 bottles of each 

114. From the middle of the side of a square 10-acre field I 
run a line cutting off 3J acres: required the length of the 
line. ■ 41 rd. 

115. What shall I pay for a bond of $100, having 5 years to 
run, with interest at 6%, in order to make it an 8% invest- 
ment? $92.86. 

116. A and B shot a deer, and wishing to ascertain its 
weight, they procured a pole and poised it on a log — A on 
one end weighing 150 lb., and B on the other weighing 180 lb., 
after which A took the deer and exchanged places with B, 
when the pole again balanced: find the weight of the deer? 

66 1b. 

117. A carpenter has to put rafters, 13 ft. long and 6 in. thick, 
upon a frame 24 ft. wide. He does not want to cut the edge 
of the frame, and yet must have just 13 ft. from their edges to 
the angle of the roof. How long must be the under side of 
each rafter? 11 ft. 7.1 in. 



250 FAIRCHILD'S SOLUTION BOOK. 

118. A teacher had his wages increased each term by 33J% 
for three succeeding terms of eight months; he now gets $64 
per month: find his salary per month the first term. $27. 

119. A broker sold a farm for $520, charging a certain rate of 
commission, and invested the proceeds, less his charges of 
4% on purchase price, in city property: if his entire commis- 
sion was $450, find his charges on sale of farm? 86%. 

120. The aggregate face value of two notes is $108, and each 
has two years to run; I have both discounted at 8%, one by 
bank discount and the other by true discount, realizing $92: 
find face of each note. $50 and $58. 

121. Alta and Marie bought a basket of oranges at the rate 
of 3 for 2 ct., and gained 50 ct. by selling them at the rate of 
2 for 3 ct.: how many oranges did they buy? 60. 

122. Invested $270 in stock at 25% discount, which pays 8 
% annual dividends: how much must I invest in stock at 4% 
discount and paying 10% annual dividends, to secure an equal 
income? $2764.80. 

123. Bought some turkeys for $26; had I bought 8 more at 
10 ct. less each, all would have cost $35.60: how many did I 
buy? 20. 

124. One agent sold a house for $3500 at 1%; another sold 
a farm for $100 more, and sent me $45 less than the first: 
what rate did the second charge? 5%. 

125. What sum of money must be invested in 4% bonds, 
selling at 125, to give an annual income of $2500? What rate 
of interest is received on the investment? $78125, 3.2%. 

126. A man bought a farm for $6000, and agreed to pay 
principal and interest in three equal annual installments: 
what was the annual payment, interest being 6%? 

$2244.658. 

127. If the time past noon is \ of the time to midnight, 
what is the hour? 4 p. m. 

128. If a man weighs 160 lb. avoirdupois, what will he weigh 
by Troy weight? 194f lb. 

129. Duck island is 73 mi. in circumference. If two men 
set out from the same point in the same direction, the one 



PROBLEMS. 251 

traveling at the rate of 5 mi. an hour and the other at the rate 
of 3 mi. an hour, in what time will they again be together? 

36^ hr. 

130. Charles Draper had a 6% bond of $800, dated Jan. 1, 
1898, due Jan. 1, 1899. On July 1, 1898, he sold the bond to 
G. K. Joseph in such a way as to give him 8% on his invest- 
ment. If Mr. Joseph borrowed the money needed to pay the 
note, from the bank at 10% for 90 days, what was the face of 
the bank note? $837,007. 

131. The width of a rectangular field is f of the length, and 
its diagonal measures 30 rd: find its area. 432 sq. rd. 

132. What is the length of a cylinder 6 in. in diameter, 
whose solidity is the same as a 12-inch sphere? 32 in. 

133. A walks 9 hr. a day and B 8 hr. a day; B's rate is 20% 
faster than A's: how long will it take A to descend a hill 
which B descends in 5 days, provided a man's rate is in- 
creased 33|% in going down hill and decreased 25% up hill? 

27 hr., or 3 da. 

134. Find the volume of a pyramid cut from a cube 4 in. 
from a corner on each edge. lOf cu. in. 

135. If I buy uncurrent bank notes at 10% discount, brok-« 
erage 2j%, and sell them at par, thus gaining $348.75, what 
is the face value? $4500. 

136. Two poles 80 ft. and 120 ft. high, respectively, are 
planted on the ground 100 ft. apart: if lines be drawn from 
the tops of the poles to the opposite bottoms, what is the dis- 
tance from the ground to where the lines cross each other? 

48 ft. 

137. $429.30. Cleveland, O., April 13, 1873. 
On demand, I promise to pay W. Jones, or order, four 

hundred and twenty-nine T 3 ° dollars. Value received, with 
interest at 6%. R. Smith. 

Indorsed: Oct. 2, 1873, $10; Dec. 8, 1873, $ ; July 17, 1874, $200. 

The exact sum due Jan. 1, 1875, was $195. Restore the lost 
figures of the second payment. $59.97. 

138. An apple is suspended by a string from the ceiling 
until it comes within 4 ft. of the floor; it is swung back 6 ft., 



252 FAIRCHILD'S SOLUTION BOOK. 

when it is 8 ft. from the floor: what is the height of the ceil- 
ing? Vd\ ft. 

139. The area of an equilateral triangle whose base falls on 
the diameter and its vertex in the middle of the arc of a semi- 
circle, is equal to 100: what is the diameter of the semicircle? 

26.3214. 

140. Find the volume of a rectangular piece of ice 10 in. 
long, 6 inches wide, and floating in water, with \ inch of its 
thickness above water, specific gravity being .9. 300 cu. in. 

141. Goliath of Gath weighed 1015 lb: what was his height, 
if a man 5 ft. 10 in. in height and of similar proportions 
weighs 180 lb.? 10 ft. 4.6 in. 

142. A house which cost $5400 rents for $30 per month: 
what is the rate per annum of interest received on the invest- 
ment if the average annual expenses are $144? 4%. 



CHAPTER XIV. 

RULES OF MENSURATION. 

RHOMBUS. 

1. Area=half the product of its diagonals. 

2. The sides intersect at right angles. 

3. Side = V( sum °f i the diagonals squared). 

CUBE. 

1. Volume=(edge) 3 . 



2. Diagonal=V(edge 2 x3), or V(area-^2). 

3. Area = edge 2 x6, or diagonal 2 X 2. 

4. All cubes are similar solids. 

PYRAMID. 

1. Volume = area of baseX-^ of the altitude. 

2. Lateral area=perimeterX^ slant height. 

FRUSTUM OF PYRAMID. 

1. Volume is found by a well known rule. 

2. Lateral area=half the sum of basal perimeter X slant 
height. 

CIRCLE. 

1. Area = ^ 2 X7r, d*X\ir. 



2. Circum.= V(areaX7r)x2. 



3. Radius^ V (area-Hr). 

4. All circles are similar. 

CYLINDER. 

1. Volume = area of baseXthe altitude. 

2. Curved surface=circumference of baseX altitude. 
(253) 



254 FAIR CHILD'S SOLUTION BOOK. 

SPHERE. 

1. Volume =^ 3 X|tt, or .5236. 

2. Area— d*Xir. 



3. Diameter^i^.vol. X^tt), or (area-Hr). 

CONE. 

1. Volume = area of base X J of altitude. 

2. Curved surface = circumference of base X -J- slant height. 

3. Altitude = V (slant height 2 — radius 2 ). 

MISCELLANEOUS. 

1. Diameter X.7071=side of inscribed square. 

2. AreaX.6366=side of inscribed square. 

3. To find the distance a nail-head in a revolving wheel 
moves, multiply the distance traveled by 4 and divide by ir. 



POINTS FOR THE STUDENT. 

106. Is the quotient always an abstract number? 

Show this in the following: Divide 10 apples equally 
among 5 boys. The quotient is not always an abstract num- 
ber. It is only true when the dividend and divisor are both 
of the same denomination. When the divisor is abstract, the 
quotient is the same as the dividend. Hence, 10 apples-f-5 = 
2 apples. And the world of to-day says, $20-^2=$10. 

107. Take a board 8 in. square, and it will contain 64 sq. in.; 
now saw it so that it will make a rectangle 5X13, and it will 
contain 65 sq. in. Also, a board 10 in. square, saw it and 
place it in a rectangular shape 6x16, and it will have but 96 
sq. in. Explain the paradoxes. 

The fallacy in these propositions consists in the fact that 
when the square of 8 in. is cut as indicated, the two parts will 
not make a rectangle, but a figure whose sides are not exactly 
straight; that is, the sides are not rightly inline. To makethe 
sides straight as inferred in the rectangle 5x13, there would 
be a gap in the cut sides, equal to a square inch. Similarly in 



JOURNALS. 



255 



the 10-inch square, the cut sides would overlap to make a true 
rectangle, and, therefore, embrace a surface of 4 sq. in. less. 

(School Visitor.) 



JOURNALS. 

The Ohio Teacher: Edited and published by John Mc- 
Burney, Cambridge, O.; published monthly; price, 75 cents 
per year. This is one of the best educational journals in 
the state. It has rendered an untold amount of good to 
hundreds of teachers. Every energetic teacher should have 
this journal on his desk. 

The American Mathematical Monthly: Edited by B. F. 
Finkel, A. M., and J. M. Colaw, A. M., Springfield, Mo.; price, 
$2.00 per year. This journal is devoted entirely to mathe- 
matics. The best mathematicians in the world contribute to 
its pages. It ranks as one of the foremost now published. 




MAY 26)899 



